Home » HC-Verma Solutions for Class 12 » HC Verma Class XII Science Physics Chapter 12 – Simple Harmonic Motion

HC Verma Class XII Science Physics Chapter 12 – Simple Harmonic Motion


Exercise : Solution of Questions on page Number : 250


Answer: 7

The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stop watch to measure the time between consecutive passage, we are certain about the mean position.


Answer: 8

Figure (a) shows the graph of the applied force against the position of the particle.


Answer: 9

No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at mean position is zero. The potential energy increases in positive direction at the extreme position.

However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.


Answer: 10

Statement A is more appropriate because the energy of a system in simple harmonic motion is given by E=12m ω2A2.

If the mass (m) and angular frequency (ω) are made constant, Energy (E) becomes proportional to the square of amplitude (A2).
i.e. ∝ A2

Therefore, according to the relation, energy increases as the amplitude increases.


Answer: 11

According to the relation:
T=2πlg

The time period (T) of the pendulum becomes proportional to the square root of inverse of g if the length of the pendulum is kept constant.
i.e. T∝1g

Also, acceleration due to gravity (g) at the poles is more than that at equator. Therefore, the time period decreases and the clock gains time.


Answer: 12

No. According to the relation:
T=2πlg
The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free falling body and its geffective (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.


Answer: 13

The time period of a pendulum depends on the length and is given by the relation, T=2πlg.
As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.


Answer: 14

Yes.

Time period of a spring mass system is given by,
T=2πmk  …(1)              .
where m is mass of the block, and
k  is the spring constant

Time period is also given by the relation,
T=2πx0g  …(2)
where, x0 is extension of the spring, and
is acceleration due to gravity

From the equations (1) and (2), we have:
mg=kx0

⇒k=mgx0

Substituting the value of k in the above equation, we get:
T=2πmmgx0=2πx0g
Thus, we can find the time period if the value of extension x0 is known.


Answer: 15

When the frequency of soldiers’ feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers’ feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers’ feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.


Answer: 16

As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.


Answer: 1

(a) As x increases k increases.

A body is said to be in simple harmonic motion only when,
   F = – kx  …(1)
where F is force,
  k is force constant, and
     x is displacement of the body from the mean position.

Given:
  F = –kx   …(2)

On comparing the equations (1) and (2), it can be said that in order to execute simple harmonic motion, k should be proportional to x . Thus, as x increases k increases.


Answer: 2

(b) an extreme position

One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.


Answer: 3

(a) vmax

Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation.


Answer: 4

(d) zero

Displacement is defined as the distance between the starting and the end point through a straight line. In one complete oscillation, the net displacement is zero as the particle returns to its initial position.


Answer: 5

(c) 4A

In an oscillation, the particle goes from one extreme position to other extreme position that lies on the other side of mean position and then returns back to the initial extreme position. Thus, total distance moved by particle is,
2A + 2A = 4A.


Answer: 6

(d) zero

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
Aω2 + ( –Aω2) = 0


Answer: 7

(d) simple harmonic with amplitude  A2+B2

x = A sin ωt + B cos ωt   …(1)

Acceleration, a=d2xdt2=d2dt2(A sin ωt + B cos ωt)   =ddt(Aω cos ωt – Bω sin ωt) =-Aω2 sinωt – Bω2 cos ωt =-ω2(A sin ωt + B cos ωt )= -ω2x

For a body to undergo simple harmonic motion,
acceleration, a = – kx.  …(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude,  A =A2+B2.2


Answer: 8

(c) on a circle

We know, d2 dt2r→=-ω2 r→
But there is a phase difference of 90o between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.


Answer: 9

(b) B

At t = 0,
Displacement x0 is given by,
x0 = A + sin ω(0) = A

Displacement x will be maximum when sinωt is 1
or,
xm = A + B

Amplitude will be:
xm  xo = A + B – A = B


Answer: 10

(c) phase

Because the direction of motion of particles A and B is just opposite to each other.


Exercise : Solution of Questions on page Number : 249


Answer: 1

No. As motion is a change in position of an object with respect to time or a reference point, it is not an example of periodic motion.


Answer: 2

No. The resultant force on the particle is maximum at the extreme positions.


Answer: 3

Yes. Simple harmonic motion can take place in a non-inertial frame. However, the ratio of the force applied to the displacement cannot be constant because a non-inertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.


Answer: 4

No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.


Answer: 5

Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.


Answer: 6

No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.


Exercise : Solution of Questions on page Number : 251


Answer: 11

 (d) remain E
Mechanical energy (E) of a spring-mass system in simple harmonic motion is given by,
E=12mω2A2
where m is mass of body, and
ω is angular frequency.

Let m1 be the mass of the other particle and ω1 be its angular frequency.
New angular frequency ω1 is given by,
ω1=km1=k2m
(m1=2m)
New energy E1 is given as,
E1 = 12m1ω12A2  =12(2m)(k2m)2A2  =12mω2A2 =E


 Answer: 12

(a)12mω2A2

It is the total energy in simple harmonic motion in one time period.


Answer: 13

(c) 2v

Because in one complete oscillation, the kinetic energy changes its value from zero to maximum, twice.


Answer: 14

 (d) become T/2T/2

Time period T is given by,
T = 2πmk
where m is the mass, and
k is spring constant.

When the spring is divided into two parts, the new spring constant k1 is given as,
k1 = 2k

New time period T:
T1  = 2πm2k=122πmk=12T


Answer: 15

(d)k2k1

Maximum velocity, v = 
where A is amplitude and ω is the angular frequency.

Further, ω = km
Let A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,

i.e.  vA=vBor, AωA=BωB⇒ Ak1mA= Bk2mB⇒   Ak1m= Bk2m (mA=mB=m)⇒ AB= k2k1


Answer: 16

(c) remain same

Because the frequency (ν=12πkm) of the system is independent of the acceleration of the system.


Answer: 17

(c) remain same

As the frequency of the system is independent of the acceleration of the system.


Answer: 18

(d) 6 times slower

The acceleration due to gravity at moon is g/6.

Time period of pendulum is given by, T = 2πlg

Therefore, on moon, time period will be:

Tmoon = 2πlgmoon=2πl(g6)=6(2πlg)=6T


Answer: 19

(d) give correct time

Because the time period of a spring-mass system does not depend on the acceleration due to gravity.


Answer: 20

(c) its length should be decreased to keep correct time

Time period of pendulum,
T = 2πlg
At higher altitudes, the value of acceleration due to gravity decreases. Therefore, the length of the pendulum should be decreased to compensate for the decrease in the value of acceleration due to gravity.


Answer: 21

(c) will go in a circular path

As the acceleration due to gravity acting on the bob of pendulum, due to free fall gives a torque to the pendulum, the bob goes in a circular path.


Answer: 1

(a) A simple harmonic motion is necessarily periodic.
(b) A simple harmonic motion is necessarily oscillatory.

A periodic motion need not be necessarily oscillatory. For example, the moon revolving around the earth.
Also, an oscillatory motion need not be necessarily periodic. For example, damped harmonic motion.


Answer: 2

(a) periodic

Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position.


Answer: 3

(a) periodic

Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position.


Answer: 4

(d) none of them

As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position.


Answer: 5

(a) periodic
(b) oscillatory
(d) angular simple harmonic

Because it completes one oscillation in a fixed interval of time and the oscillations are in terms of rotation of the body through some angle.


Answer: 6

(c) a⇀.r⇀(d) F⇀.r⇀

In S.H.M.,
F = –kx
Therefore, F.⇀r⇀ will always be negative. As acceleration has the same direction as the force, a⇀.r⇀will also be negative, always.


Answer: 7

(a) F⇀.a⇀

As the direction of force and acceleration are always same, F⇀.a⇀  is always positive.


Answer: 8

(a)F⇀ x a⇀(b)v⇀ x r⇀⇀(c)a⇀x r⇀(d)F⇀ x r⇀

As F⇀, a⇀, r⇀ and v⇀ are either parallel or anti-parallel to each other, their cross products will always be zero.


Answer: 9

(c) on a straight line
(d) periodic

If the particle were dropped from the surface of the earth, the motion of the particle would be SHM. But when it is dropped from a height h, the motion of the particle is not SHM because there is no horizontal velocity imparted. In that case, the motion of the particle would be periodic and in a straight line.


Exercise : Solution of Questions on page Number : 252


Answer: 1

It is given,
Amplitude of the simple harmonic motion, A =10 cm

At t = 0 and  x = 5 cm,
Time period of the simple harmonic motion, T  = 6 s

Angular frequency (ω) is given by,
ω=2πT=2π6=π3 sec-1

Consider the equation of motion of S.H.M,
Y = Asin ωt+ϕ                                 …(1)
where Y is displacement of the particle, and
ϕ is phase of the particle.

On substituting the values of A, t and ω in equation (1), we get:
5 = 10sin(ω × 0 + ϕ)
⇒5 = 10sin ϕ

sin ϕ=12⇒ϕ=π6

 Equation of displacement can be written as,
x=10 cm sin π3t+π6

(ii) At t = 4 s,
x=10sinπ34+π6 =10sin8π+π6 =10sin9π6 =10sin3π2 = 10sinπ+π2 =-10sinπ2=-10

Acceleration is given by,
    a = −ω2x
=-π29×-10=10.9≈11 cm/sec-2


Answer: 2

It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms−1.
Acceleration of the particle, a = 10 ms−2.

Let ω be the angular frequency of the particle.
The acceleration of the particle is given by,
     a = ω2x
⇒ω=ax=100.02  =500=105 Hz
Time period of the motion is given as, T=2πω=2π105  =2×3.1410×2.236=0.28 s

Now, the amplitude A is calculated as,
v=ωA2-x2 ⇒v2=ω2 A2-x2 1=500A2-0.0004  ⇒A=0.0489=0.049 m ⇒A= 4.9 cm


Answer: 3

It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm

To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
ω be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.

Equating the mathematical expressions for K.E. and P.E. of the particle, we get:
12mω2 A2-y2=12mω2y2
    A2 − y2 = y2
2y2 = A2
⇒ y=A2=102=52

The kinetic energy and potential energy of the particle are equal at a distance of 52 cm from the mean position.


Answer: 4

It is given that:
Maximum speed of the particle, vMax = 10 cms-1
Maximum acceleration of the particle, aMax = 50 cms−2

The maximum velocity of a particle executing simple harmonic motion is given by,
vMax=Aω
where ω is angular frequency, and
is amplitude of the particle.

Substituting the value of vMax in the above expression, we get:
 = 10   …(1)
⇒ω2=100A2

   aMax = ω2A = 50 cms−1
⇒ω2=50A  …(2)From the equations (1) and (2),
we get : 100A2=50A ⇒A=2 cm
∴  ω=100A2=5 sec-1

To determine the positions where the speed of the particle is 8 ms-1, we may use the following formula:
      v2 = ω2 (A2 − y2)
where y is distance of particle from the mean position, and
v is velocity of the particle.

On substituting the given values in the above equation, we get:
64 = 25 (4 − y2)
⇒6425=4-y2

⇒  4 − y2 = 2.56
⇒   y2 = 1.44
⇒​   y  = 1.44
⇒  y = ± 1.2 cm   (from the mean position)


Answer: 5

Given:
Equation of motion of the particle executing S.H.M.,
x=2.0 cm sin 100 s-1t+π6Mass of the particle, m=10 g   …(1)
General equation of the particle is given by,
x = Asin(ωt+ϕ)   …(2)

On comparing the equations (1) and (2) we get:

(a) Amplitude, A is 2 cm.
Angular frequency, ω is 100 s−1​.

Time period is calculated as,  T=2πω=2π100=π50s   =0.063 s

Also, we know –
T=2πmk where k is the spring constant.
⇒T2=4π2mk⇒  k=4π2mT2=105 dyne/cm  =100 N/m

(b) At t = 0 and x = 2 cm sinπ6
=2×12=1 cm from the mean position,

We know:
     x = A sin (ωt + ϕ)

Using v=dxdt, we get :
    v = Aω cos (ωt + ϕ)
=2×100 cos 0+π6=200×32=1003 cms-1=1.73 ms-1

(c) Acceleration of the particle is given by,
    a = -ω2x
= 1002×1 = 10000 cm/s2


Answer: 6

Given:
The equation of motion of a particle executing S.H.M. is,
x=5 sin 20t+π3

The general equation of S..H.M. is give by,
x=A sin (ωt+ϕ)

(a) Maximum displacement from the mean position is equal to the amplitude of the particle.
As the velocity of the particle is zero at extreme position, it is at rest.
    ∴  Displacement x = 5, which is also the amplitude of the particle.
⇒  5=5 sin 20t+π3Now,   sin 20t+π3=1=sinπ2   ⇒  20t+π3=π2   ⇒  t=π120 s

The particle will come to rest at π120 s

(b)  Acceleration is given as,
a = ω2x
=ω25 sin 20t+π3

For a = 0,
5 sin 20t+π3=0⇒sin 20t+π3=sin π⇒20t=π-π3=2π3⇒    t=π30 s

(c) The maximum speed v is given by,
v=Aωcos ωt+π3  (using v=dxdt)
=20×5 cos 20t+π3
For maximum velocity:
cos 20t+π3=-1= cos π⇒20t=π-π3=2π3⇒   t=π30 s


Answer: 7

It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50πt + tan−10.75)
= 2.0 cos (50πt + 0.643)

(a) Velocity of the particle is given by,
v=dxdt
     v = −100π sin (50πt + 0.643)

As the particle comes to rest, its velocity becomes be zero.
⇒​ v = −100π sin (50πt + 0.643) = 0
⇒                   sin (50πt + 0.643) = 0 = sin π

When the particle initially comes to rest,
50πt + 0.643 = π
⇒   t = 1.6 × 10−2 s

(b) Acceleration is given by,
a=dvdt =-100π×50π cos 50πt+0.643

For maximum acceleration:
cos (50πt + 0.643) = −1 = cos π (max) (so that a is max)
⇒   t = 1.6 ×  10−2 s

(c) When the particle comes to rest for the second time, the time is given as,
50πt + 0.643 = 2π
⇒ ​  t = 3.6 × 10−2 s


Answer: 8

As per the conditions given in the question,
y1=A2;  y2=A  (for the given two positions)

Let y1 and y2 be the displacements at the two positions and be the amplitude.
Equation of motion for the displacement at the first position is given by,
y1 = Asinωt1
As displacement is equal to the half of the amplitude,
A2=A sin ωt1
⇒     sin ωt1=12⇒2π×t1T=π6⇒  t1=T12

The displacement at second position is given by,
 y2 = A sin ωt2
As displacement is equal to the amplitude at this position,
⇒    A = A sin ωt2
⇒ sinωt2 = 1
⇒  ωt2=π2⇒2πTt2=π2
∵ sin π2=1⇒  t2=T4
∴ t2-t1=T4-T12=T6


Answer: 9

Given:
Spring constant, =0.1 N/m
Time period of the pendulum of clock, T = 2 s
Mass attached to the string, m, is to be found.

The relation between time period and spring constant is given as,
T=2π mk
On substituting the respective values, we get:
2 =2πmk⇒ π2m0.1=1
∴ m=0.1π2=0.110 =0.01 kg≈10 g


Answer: 10

An equivalent simple pendulum has same time period as that of the spring mass system.
The time period of a simple pendulum is given by,
Tp=2πlg
where l is the length of the pendulum, and
g is acceleration due to gravity.

Time period of the spring is given by,
Ts=2πmk
where is the mass, and
is the spring constant.

Let x be the extension of the spring.
For small frequency, TP ​can be taken as equal to TS.


⇒lg=mk⇒lg=mk⇒l=mgk=Fk=x

(∵ restoring force = weight = mg)

∴  l = x (proved)


Answer: 11

It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T  = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5×10 = 5 kg
∵ g=10 ms-2

Total force exerted on the block = Weight of the block + spring force

Periodic time of spring is given by,
T=2πmk⇒0.314=2π0.5k⇒      k=200 N/m

∴ The force exerted by the spring on the block F is,
F = kx = 200.0 × 0.1 = 20 N

Maximum force = + weight of the block
= 20 + 5 = 25 N


Answer: 12

It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,
T=2πmk
where k is the spring constant.

On substituting the respective values, we get:
⇒4=2π2k⇒2=π2k⇒4=π22k⇒k=2π24  =π22= 5 N/m

As the restoring force is balanced by the weight, we can write:
mg kx
⇒ x=mgk=2×105=4

∴ Potential Energy U of the spring is,
U=12kx2=12×5×16  =5×8=40 J.


 Answer: 10

(a) displacement from the mean position

For S.H.M.,
F = -kx
ma = – kx   (F = ma)
or
= -kmx
Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.


Answer: 11

(a) on a straight line
(c) periodic
(d) simple harmonic

The given equation is a solution to the equation of simple harmonic motion.
The amplitude is (i⇀+2j⇀)A, following equation of straight line mx c.
Also, a simple harmonic motion is periodic.


Answer: 12

(d) with time period πω

Given equation:
x = xo sin2 ωt

⇒​ x=x02(cos 2ωt- 1)

Now, the amplitude of the particle is xo/2 and the angular frequency of the SHM is 2ω.

Thus, time period of the SHM = 2πangular frequency=2π2ω=πω


Answer: 13

(d) the average potential energy in one time period is equal to the average kinetic energy in this period.

The kinetic energy of the motion is given as,
12kA2 cos2 ωt

The potential energy is calculated as,
12kA2 sin2 ωt

As the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.


Answer: 14

(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy

In SHM,
maximum kinetic energy    = 12kA2
maximum potential energy = 12kA2

The minimum value of both kinetic and potential energy is zero.
Therefore, in a simple harmonic motion the maximum kinetic energy and maximum potential energy are equal. Also, the minimum kinetic energy and the minimum potential energy are equal.


Answer: 15

(a) the measured times are same
(b) the measured speeds are same

The effect of gravity on the object as well as on the pendulum clock is same in both cases; the time measured is also same. As the time measured is same, the speeds are same.


Answer: 16

(a) A simple pendulum
(b) A physical pendulum

As the time period of a simple pendulum and a physical pendulum depends on the acceleration due the gravity, the time period of these pendulums changes when they are taken to the moon.


Exercise : Solution of Questions on page Number : 253


Answer: 16

It is given that:
Spring constant, k = 100 N/m,
Mass of the block, M = 1 kg
Force, F = 10 N

(a) In the equilibrium position,
kx
where x is the compression of the spring, and
k is the spring constant.

 F10100 =0.1 m=10 cm

(b) The blow imparts a speed of 2 ms-1 to the block, towards left.
Potential energy of spring, U = 12kx2
Kinetic energy, K = 12Mv2

(c) Time period (T) is given by,
T = 2πMk
=2π1100=π5 s

(d) Let A be the amplitude.
Amplitude is the distance between the mean and the extreme position.
At the extreme position, compression of the spring will be (A + x).

As the total energy in S.H.M.  remains constant, we can write:
12k(A+x)2=12kx2+12Mv2+Fx  =2.5+10x             

 50(A + 0.1)2 = 2.5 + 10x
 50A2 + 0.5 + 10A = 2.5 + 10A
 50A2 = 2
⇒ A2=250=4100A=210 m=0.2 m=20 cm

(e) Potential Energy at the left extreme will be,
P.E.=12k(A+x)2=12×100×(0.1+0.2)2 =50×(0.09)=4.5 J

(f) Potential Energy at the right extreme is calculated as:

Distance between the two extremes = 2A
P.E.12(A+x)− F(2A)
= 4.5 − 10 (0.4) = 0.5 J

As the work is done by the external force of 10 N, different values of options (b), (e) and (f) do not violate the law of conservation of energy.


Answer: 17

(a) Spring constant of a parallel combination of springs is given as,
K = k1 + k2  (parallel)
Using the relation of time period for S.H.M. for the given spring-mass system, we have:
T=2πmK=2πmk1+k2

(b) Let be the displacement of the block of mass m, towards left.
Resultant force is calculated as,
F = F1 + F2 = (k1 + k2)x

Acceleration (a) is given by,
a=(Fm)=(k1+k2)mx

Time period (T) is given by,
T=2πdisplacement accelerationOn substituting the values of displacement and acceleration,
we get T =2πxx(k1+k2)m =2πmk1+k

Required spring constant, K = k1 + k2

(c) Let K be the equivalent spring constant of the series combination.
 1K=1k1+1k2=k2+k1k1k2K=k1k2k1+k2
Time period is given by,T=2πmKOn substituting the respective values,
we get T=2πm(k1+k2)k1k2


Answer: 18

(a) We know-
f = kx
x=FkAcceleration=Fm

Using the relation of time period of S.H.M.,

.Time period, T=2πDisplacement Acceleration   =2π(Fk)(Fm)=2πmk

Amplitude = Maximum displacement Fk

When the block passes through the equilibrium position, the energy contained by the spring is given by,
12kx12(Fk)1(F2k)

(b) At the mean position, potential energy is zero.
Kinetic energy is given by,
12kx12F2k


Answer: 19

(a) Let us push the particle lightly against the spring C through displacement x.

As a result of this movement, the resultant force on the particle is kx​.
The force on the particle due to springs A and B is kx2.
Total Resultant force kx+(kx2)2+(kx2)2kx + kx = 2kx
Acceleration is given by kxm
Time period=2πDisplacement Acceleration =2πx2kx/m√ 2πm2k


Answer: 20

As the particle is pushed against the spring C by the distance x, it experiences a force of magnitude kx.

If the angle between each pair of the springs is 120˚ then the net force applied by the springs A and B is given as,
(kx2)2+(kx2)2+2(kx2)(kx2) cos 120° =kx2
Total resultant force(F) acting on mass m will be,
F=kx+kx2=3kx2           
 ∴  a=Fm=3kx2max=3k2m=ω2ω=3k2m
∴  Time period, T=2πω=2π2m3k


Answer: 21

As the block of mass is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.

Now, springs k2 and k3 are in connected as a series combination.
Let k4 be the equivalent spring constant.
∴ 1k4=1k2+1k3=k2+k3k2k3k4=k2k3k2+k3

 k4 and k1 form a parallel combination of springs. Hence, equivalent spring constant k = k1 + k4.
k2k3k2+k3+k1=k2k3+k1k2+k1k3k2+k3
∴ Time peiod, T=2πMk=2πM (k2+k3)k2k3+k1k2+k1k3

(b) Frequency(v) is given by,
 v=1T
 = 12πk2k3+k1k2+k1k3M(k2+k3)

(c) Amplitude ( ) is given by,
 x=Fk=F(k2+k3)k1k2+k2k3+k1k3


Answer: 22

All three spring attached to the mass M are in series.
k1k2k3 are the spring constants.
Let be the resultant spring constant.
1k=1k1+1k2+1k3k=k1k2k3k1k2+k2k3+k3k1
Time period (T) is given by,T=2πM =2M(k1k2+k2k3+k3k1)k1k2k3 =2M(1k1+1k2+1k3)

As force is equal to the weight of the body,
F = weight = Mg
Let x1x2, and x3 be the displacements of the springs having spring constants k1k2 and k3respectively.
​For spring k1,
 x1=MgkSimilarly, x2=Mgk2and x3=Mgk3
∴  PE1=12k1x21  =12k1(Mgk1)2  =12k1M2g2k21 =12M2g2k1=M2g22k1
Similarly, PE2=M2g22kand PE3=M2g22k3


Answer: 13

It is given that:
Energy stored in the spring, E = 5 J
Frequency of the mass-spring system, = 5
Extension in the length of the spring, = 25 cm = 0.25 m
Time period, T =15 Potential energy(U) is given by, U=12kx212kx2=512k(0.25)2=5k=160 N/mTime period of spring mass system is given by, T=2π(mk) where m is the mass of the body hanged, and  k is the spring constant.On substituting the respective values in the above expression,
we get : 15=2π(m160)m=0.16 kg


Answer: 14

(a) Consider the free body diagram.
Weight of the body, W = mg
Force, F = ma = mω2x

x is the small displacement of mass m.
As normal reaction is acting vertically in the upward direction, we can write:
R + mω2x − mg = 0 ….(1)

Resultant force = mω2x = mg − R
⇒ mω2x=m(kM+m)x =mkxM+mHereω={kM+m}        

(b) R = mg − mω2x
=mgmkM+Nx=mgmkxM+N
It can be seen from the above equations that, for R to be smallest, the value of mω2x should be maximum which is only possible when the particle is at the highest point.

(c) R = mg − mω2x
As the two blocks oscillate together becomes greater than zero.
When limiting condition follows,
 i.e. R = 0
      mg = mω2x
x=mgmω2=mg·(M+m)mk

Required maximum amplitude=g(M+m)k


Answer: 15

(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m1 + m2sin θ

At equilibrium,
  kx = (m1 + m2sin θ
x=(m1+m2) g sin θk

(b) It is given that:
Distance at which the spring is pushed, x1=2k(m1+m2)g sin θ  

As the system is released, it executes S.H.M.
where ω=km1+m2

When the blocks lose contact, becomes zero. (is the force exerted by mass m1 on mass m2)
∴  m2g sin θ=m2x2ω2=m2x2×km1+m2x2=(m1+m2) g sin θk

Therefore, the blocks lose contact with each other when the spring attains its natural length.

(c) Let v be the common speed attained by both the blocks.


Exercise : Solution of Questions on page Number : 254


Answer : 23

Let l be the extension in the spring when mass m is hung.


Let T1 be the tension in the string; its value is given by,
T1 = kl = mg 
Let x be the extension in the string on applying a force F.
Then, the new value of tension T2 is given by,
T2 = k(x + l)
Driving force is the difference between tensions T1 and T2.
∴ Driving force = T2 − T1 = k(x + l) − kl
kx
Acceleration, a=kxmTime period T is given by,T=2π displacement Acceleration  =2πxkx/m=2πmk


Answer: 24

Let us try to solve the problem using energy method.
If δ is the displacement from the mean position then, the initial extension of the spring from the mean position is given by,
δ = mgk
Let x be any position below the equilibrium during oscillation.
Let v be the velocity of mass m and ω be the angular velocity of the pulley.
If r is the radius of the pulley then
v = rω
As total energy remains constant for simple harmonic motion, we can write:

12Mv2+12Iω2+12k(x+δ)2-δ2-Mgx=Constant⇒12Mv2+12Iω2+12kx2+kxd-Mgx=Constant⇒12Mv2+12Iv2r2+12kx2=Constant      ∵δ=Mgk
By taking derivatives with respect to t, on both sides, we have:

Mv.dvdt+Ir2v.dvdt+kxdxdt=0Mva+Ir2va+kxv=0
∵v=dxdt and a=dvdtaM+Ir2=-kx ⇒ax=kM+Ir2=ω2T=2πω⇒T=2πM+Ir2k


Answer: 25


The centre of mass of the system should not change during simple harmonic motion.
Therefore, if the block m on the left hand side moves towards right by distance x, the block on the right hand side should also move towards left by distance x. The total compression of the spring is 2x.
If v is the velocity of the block. Then
Using energy method, we can write:
12k2x2+12mv2+12mv2=C
⇒ mv2 + 2kx2 = C
By taking the derivative of both sides with respect to t, we get:
2mvdvdt+2k×2xdxdt=0Putting v=dxdt;and a=dvdtin above expression, we get ma+2kx=0 ⇒-ax=2km=ω2⇒ω=2km⇒Time period, T=2πm2k


Answer: 26

Let m is the mass of rectangular plate and x is the displacement of the rectangular plate.
During the oscillation, the centre of mass does not change.
Driving forceF is given as,
F = mgsin θ
Comparing the above equation with F = ma, we get:
a = Fm = gsinθ
For small values of θ, sinθ can be taken as equal to θ.
Thus, the above equation reduces to:
a=gθ=gxL  Where g and L are constant.
It can be seen from the above equation that, a α x.
Hence, the motion is simple harmonic.
Time period of simple harmonic motion T is given by,
T=2π displacement Acceleration  =2πxgx/L=2πLg


Answer: 27

It is given that:
Amplitude of simple harmonic motion, x  = 0.1 m
Total mass of the system, M = 3 + 1 = 4 kg(when both the blocks move together)
Spring constant, = 100 N/m
​Time period of SHM T is given by,
T=2πMkOn substituting the values of M and k in the bove equation,
we have : T=2π4100=2π5 s Frequency of the motion is given by, 1T=52π Hz
Let v be the velocity of the 1 kg block, at mean position.
As kinetic energy is equal to the potential energy, we can write:12mv2=12kx2
where = amplitude = 0.1 m
Substituting the value of x in above equation and solving for v, we get:12×1×v2=12×1000.12 v=1 ms-1  …1
When the 3 kg block is gently placed on the 1 kg block, the 4 kg mass and the spring become one system. As a spring-mass system experiences external force, momentum should be conserved.
Let V be the velocity of 4 kg block.
Now,
Initial momentum = Final momentum
∴ 1 × v = 4 × V
⇒V=14 m/s  As v=1 ms-1, from equation (1)
Thus, at the mean position, two blocks have a velocity of 14ms-1.
Mean value of kinetic energy is given as,KE at mean position=12MV2 =12×4×142=12×14=18
At the extreme position, the spring-mass system has only potential energy.
PE=12kδ2=12×14
where δ is the new amplitude.
∴  14=100 δ2      =δ=1400     =0.05 m = 5 cm


Answer: 28


According to the question, the collision is elastic and the surface is frictionless, therefore, when the left block A moves with speed v and collides with the right block B, it transfers all the energy to the right block B.
The left block A moves a distance x against the spring; the right block returns to the original position and completes half of the oscillation.

Therefore, the period of right block B will be,
T =2πmk2=πmk
Right block B collides with left block A and comes to rest.
Let L be the distance moved by the block to return to its original position.
The time taken is given by,
LV+LV=2LV
Hence, time period of the periodic motion is, 2LV+πmk.


Answer: 29


Let t1 and t2  be the time taken by the particle to travel distances AB and BC respectively.
Acceleration for part AB, a1 = sin 45°
The distance travelled along AB is s1.

∴ s1=0.1sin 45°=2 m

Let v be the velocity at point B, and
u be the initial velocity.

Using the third equation of motion, we have:
v2 − u2 = 2a1s1
⇒v2=2×g sin 45°×0.1sin 45°=2⇒v=2 m/sAs v=u+a1t1
∴ t1=v-ua1 =2-0g2  =2g=210=0.2 sec  ( g=10 ms-2)

For the distance BC,
Acceleration, a2 = – gsin 60°
Initial velocity, u=2 v = 0
∴ time period, t2=0-2-g32=223g=2×1.4141.732×10=0.163 s

Thus, the total time period, t = 2(t1 + t2) = 2 (0.2 + 0.163) = 0.73 s


Answer: 30

Let x1 and x2 be the amplitudes of oscillation of masses m and M respectively.

(a) As the centre of mass should not change during the motion, we can write:
mx1 = Mx2  …(1)

Let k be the spring constant. By conservation of energy, we have:
12kx02=12kx1+x22   …(2)
where x0 is the length to which spring is stretched.

From equation (2) we have,
x0=x1+x2

On substituting the value of x2 from equation (1) in equation (2),
we get : x0=x1+mx1M⇒x0=1+mMx1⇒x1=MM+mx0

Now, x2=x0-x1
On substituting the value of x1 from above equation, we get:
⇒ x2=x01-MM+m⇒ x2=mx0M+m

Thus, the amplitude of the simple harmonic motion of a car, as seen from the road is mx0M+m.

(b) At any position,
Let v1 and v2 be the velocities.

Using law of conservation of energy we have,
12Mv2+12mv1-v22+12kx1+x22=constant  …3

Here, (v1 − v2) is the absolute velocity of mass m as seen from the road.

Now, from the principle of conservation of momentum, we have:
Mx2 = mx1

⇒ x1=Mmx2  ….4

Mv2=mv1-v2⇒v1-v2=Mmv2  ….5

Putting the above values in equation (3), we get:
12Mv22+12mM2m2v22+12kx221+Mm2=constant

∴ M1+Mmv22+k1+Mmx22=constant⇒Mv22+k1+Mmx22=constant

Taking derivative of both the sides,
we get : M×2v2dv2dt+kM+mm2x2dx2dt=0⇒ma2+kM+mmx2=0
because, v2=dx2dta2x2=-kM+mMm=ω2

∴  ω=kM+mMmTherefore, time period, T=2πMmkM+m


Answer: 31

Let x be the displacement of the uniform plate towards left.
Therefore, the centre of gravity will also be displaced through displacement x.

At the displaced position,
R1 + R2 = mg

Taking moment about g, we get:
R1L2-x=R2L2+x=Mg-R1 L2+x  …1
∴ R1L2-x=Mg-R1 L2+x⇒R1L2-R1x=MgL2-R1x+Mgx-R1L2⇒R1L2+R1L2=Mgx+L2⇒R1L2+L2=Mg2x+L2⇒R1L=Mg2x+L⇒R1=MgL+2x2LNow, F1=μR1=μMgL+2x2L
Similarly, F2=μR2=μMgL-2x2LAs F1>F2,
we can write:F1-F2=Ma=2μMgLx   ax=2μgL=ω2⇒ω=2μgL
Time period T is given by,T=2πL2μg


Answer: 32

It is given that:
Time period of the second pendulum, T = 2 s
Acceleration due to gravity of a given place, g = π2 ms−2
The relation between time period and acceleration due to gravity is given by,
T=2πlg
where l is the length of the second pendulum.

Substituting the values of T and g, we get:
⇒2=2πlπ2⇒1π=lπ⇒l=1 m

Hence, the length of the pendulum is 1 m.


Answer: 33

It is given that:Angle made by the simple pendulum with the vertical, θ=π90sin πs-1tOn comparing the above equation with the equation of S.H.M.,
we get:  ω= π s-1  ⇒2πT=π
∴ T= 2 s Time period is given by the relation,
T=2πlg⇒2=2πlπ2⇒1=π1πl⇒l=1 m
Hence, length of the pendulum is 1 m.


Exercise : Solution of Questions on page Number : 255


Answer: 46

It is given that a car is moving with speed on a circular horizontal road of radius r.
(a) Let T be the tension in the string.

According to the free body diagram, the value of is given as,
T=mg2+mv2r2
=mg2+v4r2=ma,
where acceleration, a =g2+v4r2
The time period T is given by,
T=2πlgOn substituting the respective values, we have:T=2πlg2+v4r21/2


Answer: 34

Given,
Time period of the clock pendulum = 2.04 s

The number of oscillations made by the pendulum in one day is calculated as
Nmber of seconds in one daytime period of pendulum in seconds=24×36002 = 43200

In each oscillation, the clock gets slower by (2.04 − 2.00) s, i.e., 0.04 s.
In one day, it is slowed by = 43200 × (0.04)
= 28.8 min
Thus, the clock runs 28.8 minutes slow during 24 hours.


Answer: 35

Let T1 be the time period of pendulum clock at a place where acceleration due to gravity g1 is 9.8 ms−2.
Let T1 = 2 s
g1 = 9.8 ms-2

Let T2 be the time period at the place where the pendulum clock loses 24 seconds during 24 hours.
Acceleration due to gravity at this place is g2.

T2=24×360024×3600-242  =2×36003599
As T∝1g

∴T1T2=g2g1

⇒ g2g1=T1T22⇒g2=9.8 359936002=9.795 m/s2


Answer: 36

It is given that:
Length of the pendulum, l = 5 m
Acceleration due to gravity, g = 9.8 ms-2
Acceleration due to gravity at the moon, g‘ = 1.67 ms-2

(a) Time period T is given by,
T=2πlg
=2π59.8=2π0.510=2π 0.71 s
i.e. the body will take  2π(0.7) seconds to complete an oscillation.

Now, frequency f is given by,
f=1T
∴ f=12π0.71  =0.70π Hz

(b) Let g’ be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon T’, is given as:
T’=2πlg’
On substituting the respective values in the above formula, we get:
T’=2π51.67
Therefore, frequency f’ will be,
f’=1T’   =12π1.675=12π0.577   =12π3  Hz


Answer: 37


Let the speed of bob of the pendulum at an angle θ be v.
Using the principle of conservation of energy between the mean and extreme positions, we get:
12mv2 − 0 = mgl(1 − cos θ)
 v2 = 2gl(1 − cos θ)                  …(1)

In a moving pendulum, the tension is maximum at the mean position, whereas it is minimum at the extreme position.
Maximum tension at the mean position is given by
Tmax = mg + 2mg(1 − cos θ)
Minimum tension at the extreme position is given by
Tmin = m g cosθ
According to the question,
Tmax = 2Tmin
⇒ mg + 2mg − 2m g cosθ = 2m g cosθ
⇒ 3mg = 4 mg cosθ
⇒ cos θ=34⇒θ=cos-1 34


Answer: 38


It is given that R is the radius of the concave surface.
​Let N be the normal reaction force.

Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.
∴θ=xR⇒a=gθ=gxR⇒ax=gR
⇒a=xgR
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.

Time periodT is given by,
T=2πdisplacementAcceleration
=2πxgx/R=2πRg


Answer: 39

Let ω be the angular velocity of the system about the point of suspension at any time.
Velocity of the ball rolling on a rough concave surface vC is given by,
vc = (R − r)ω
Also, vc = 1
where ω1 is the rotational velocity of the sphere.

⇒ω1=vcr=R-rrω            ⋯1

As total energy of a particle in S.H.M. remains constant,
mgR-r 1-cos θ+12mvc2+12Iω12=constantSubstituting the values of vc and ω1 in the above equation,
we get : mg R-r 1-cos θ+12mR-r2 ω2+12mr2R-rrω2=constant
∵ I = mr2mgR-r 1-cos θ+12mR-r2 ω2+15mr2 R-rrω2=constant⇒gR-r 1-cos θ+R-r2ω2 12+15=constant
Taking derivative on both sides, we get:
gR-rsinθdθdt=710R-r22ωdωdt⇒gsinθ=2×710R-rα
∵ a=dωdt⇒gsinθ=75R-rα⇒α=5gsinθ7R-r  =5gθ7R-r∴αθ=ω2=5g7R-r=constant

Therefore, the motion is S.H.M.
ω=5g7R-r Time period is given by
,⇒T=2π7R-r5g


Answer: 40

It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth’s surface, g = 9.8 ms-2

Let g’ be the acceleration due to gravity at a depth of 1600 km from the surface of the earth.
Its value is given by,
g’=g1-dR where d is the depth from the earth surface, R is the radius of earth, and  g is acceleration due to gravity. On substituting the respective values, we get: g’=9.81-16006400   =9.81-14   =9.8×34=7.35 ms-2

Time period is given as,
T=2πlg’
⇒T=2π0.47.35⇒T=2×3.14×0.23  =1.465≈1.47 s


Answer: 41

Given:
Radius of the earth is R.
Let be the total mass of the earth and ρ be the density.
Let mass of the part of earth having radius x be M‘.
∴M’M=ρ×43πx3ρ×43πR3=x3R3⇒M’=Mx3R3

Force on the particle is calculated as,
Fx=GM’mx2  =GMmR3x …1

Now, acceleration ax of mass M‘ at that position is given by,
ax=GMR3x⇒axx=ω2=GMR3=gR
∵ g=GMR2So, Time period of oscillation, T=2πRg

(a) Velocity-displacement equation in S.H.M is written as,
V=ωA2-y2  where, A is the amplitude; and  y is the displacement.

When the particle is at y = R,
The velocity of the particle is gR and  ω=gR.
On substituting these values in the velocity-displacement equation, we get:
gR=gRA2-R2    ⇒R2=A2-R2⇒A=2R

Let t1 and t2 be the time taken by the particle to reach the positions X and Y.
Now, phase of the particle at point X will be greater than π2 but less than π.
Also, the phase of the particle on reaching Y will be greater than π but less than 3π2.

Displacement-time relation is given by,
y = A sin ωt

Substituting y = R and A =2R , in the above relation, we get:
R=2R sin ωt1
⇒ωt1=3π4

Also, R=2R sin ωt2

⇒ωt2=5π4So, ωt2-t1=π2⇒t2-t1=π2ω=π2gR

Time taken by the particle to travel from X to Y:
t2-t1=π2ω=π2Rg s

(b) When the body is dropped from a height R

Using the principle of conservation of energy, we get:
Change in P.E. = Gain in K.E.
⇒GMmR-GMm2R=12mv2⇒v=gR

As the velocity is same as that at X, the body will take the same time to travel XY.

(c) The body is projected vertically upwards from the point X with a velocity gR. Its velocity becomes zero as it reaches the highest point.
The velocity of the body as it reaches X again will be,
v=gR
Hence, the body will take same time i.e. π2Rg s to travel XY.


Answer: 42

If ρ is the density of the earth, then mass of the earth M is given by,
M=43πR3ρSimilarly, mass M’ of the part of earth having radius x is given by,M’=43πx13ρM’=MR3x13

(a) Let F be the gravitational force exerted by the earth on the particle of mass m. Then, its value is given by,
F=GM’mx12Substituting the value of M’ in the above equation,
we get : F=GMmR3x13x12 =GMmR3x1=GMmR3x2+R24

(b)
Fy=Fcosθ  =GMmx1R3xx1=GMmxR3Fx=Fsin θ  =GMmx1R3R2x1=GMm2R2
(c)

Fx=GMmR2
∵ Normal force exerted by the wall N = Fx
(d)The resultant force is GMmxR3
(e) Acceleration = Driving force/mass
=GMmxR3m=GMxR3

⇒ a ∝ x (the body executes S.H.M.)
ax=ω2=GMR3⇒ω=GmR3⇒T=2πR3GM


Answer: 43

The length of the simple pendulum is l.
​Let x be the displacement of the simple pendulum..
(a)

From the diagram, the driving forces f is given by,
f = m(g + a0)sinθ                 …(1)
Acceleration (a) of the elevator is given by,
a=fm =g+a0sinθ =g+a0xl   From the diagram sinθ=xl
[ when θ is very small, sin θ → θ = x/l]

∴ a=g+a0lx  …(2)
As the acceleration is directly proportional to displacement, the pendulum executes S.H.M.
Comparing equation (2) with the expression a =ω2x,
we get :
ω2=g+a0l

Thus, time period of small oscillations when elevator is going upward(T) will be:
T=2πlg+a0

(b)

When the elevator moves downwards with acceleration a0,
Driving force (F) is given by,
F = m(g − a0)sinθ
On comparing the above equation with the expression, F = ma,
Acceleration, a=g-a0 sinθ=g-a0xl=ω2xTime period of elevator when it is moving downwardT’ is given by,T’=2πω=2πlg-a0
(c) When the elevator moves with uniform velocity, i.e. a0 = 0,
For a simple pendulum, the driving force F is given by,
F=mgxl Comparing the above equation with the expression,
F=ma, we get:a=gxl⇒xa=lgT=2π displacement Acceleration =2πlg


Answer: 44

It is given that:
Length of the simple pendulum, l = 1  feet
Time period of simple pendulum, T = π3 s
Acceleration due to gravity, g = 32 ft/s2

Let be the acceleration of the elevator while moving upwards.

Driving forcef is given by,
f = m(g + a)sinθ

Comparing the above equation with the expression, f = ma, we get:
Acceleration, a  =  (g + a)sinθ = (g +a)θ   (For small angle θ, sin θ → θ)
=g+axl=ω2x (From the diagram θ=xl)
⇒ω= g+al
Time period T is given as,
T=2πlg+a
On substituting the respective values in the above formula, we get:
π3=2π132+a19=4132+a⇒32+a=36⇒a=36-32=4 ft/s2


Answer: 45

It is given that:
When the car is moving uniformly, time period of simple pendulum, T = 4.0 s
As the accelerator is pressed, new time period of the pendulum, T’ = 3.99 s
Time period of simple pendulum, when the car is moving uniformly on a horizontal road is given by,
T=2πlg⇒4=2πlg
Let the acceleration of the car be a.
The time period of pendulum, when the car is accelerated, is given by:
T’=2πlg2+a212⇒3.99=2πlg2+a212Taking the ratio of T to T’,
we get : TT’=43.99=g2+a21/4g
On solving the above equation for a,
we get:
a=g10 ms-2


Answer: 47

Given,
Length of the long, light suspension wire, l = 3 cm = 0.03 m
​Acceleration due to gravity, g = 9.8 ms-2
(a) Time period T is given by
T = 2πlg  =2π0.039.8  = 0.34 second

(b)  Velocity of merry-go-round, v = 4 ms-1
Radius of circle, r = 2 m
As the lady sits on the merry-go-round, her earring experiences centripetal acceleration.
Centripetal acceleration a is given by,
a=v2r=422=8 m/s2
Resultant acceleration A is given by,
A=g2+a2 =96.04+64 =12.65 m/s2
Time period, T=2πlA
=2π0.0312.65 =0.30 second


Answer: 48

(a) Moment of inertia I about the point X is given by,

 I = IC.G + mh2
=ml212+mh2=ml212+m0.32=m112+0.09=m1+1.0812=m2.0812

The time period T is given by,
T=2πI mgl where I= the moment of inertia, and       l = distance between the centre of gravity and the point of suspension.On substituting the respective values in the above formula,
we get : T=2π2.08 mm×12×9.8×0.3  =1.52 s

(b) Moment of inertia I about A is given as,
   I IC.G. + mr2 = mr2 + mr2 = 2mr2

The time period (T) will be,
T=2πImglOn substituting the respective values in the above equation,
we have : T=2π2mr2mgr  =2π2rg

(c) Let I be the moment of inertia of a uniform square plate suspended through a corner.
I=ma2+a23=2m3a2

In the △ABC, l2 + l2 = a2
∴l=a2⇒T=2πImgl  =2π2ma23mgl  =2π2a2 3ga2  =2π8a3g

(d)  We know h=r2
Distance between the C.G. and suspension point, l = r2

Moment of inertia about A will be:
l = IC.G. + mh2
= mr22+mr22=mr212+14=34mr2

Time period (T) will be,
T=2πImgl =2π3mr24mgl =2π3r2g


Answer: 49

It is given that the length of the rod is l.

Let point A be the suspension point and point B be the centre of gravity.

Separation between the point of suspension and the centre of mass, l’ = l2

Also, h = l2

Using parallel axis theorem, the moment of inertia about A is given as,
I=ICG+mh2 =ml212+ml24=ml23the time period T is given by,T=2πImgl’=2πImgl2 =2π2ml23mgl=2π2l3g

Let T‘ be the time period of simple pendulum of length x.

Time period (T’) is given by,
T’=2πxgAs the time period of the simple pendulum is equal to the time period of the rod,T’=T⇒2l3g=xg⇒x=2l3


Answer: 50

Let m be the mass of the disc and r be its radius.

Consider a point at a distance x from the centre of gravity.
Thus, l = x

Moment of intertia I about the point x will be,
I = IC.G +mx2
= mr22+mx2=mr22+x2

Time period(T) is given as,
T=2πImglOn substituting the respective values in the above equation,

we get : T=2πmr22+x2mgx  (l = x)  =2πmr2+2x22mgx  =2πr2+2x22gx      …(1)

To determine the minimum value of T,
d2Tdx2=0

Now,
d2Tdx2=ddx4π2r22gx+4π22x22gx⇒2π2r2g-1×2+4π2g=0
⇒-π2r2gx2+2π2g=0⇒π2r2gx2=2π2g⇒2×2=r2⇒x=r2

Substituting this value of in equation (1),
we get : T=2πr2+2r222gx  =2π2r22gx=2πr2gr2    =2π2r2gr==2π2rg


Answer: 51

It is given that:
Radius of the hollow sphere, r = 2 cm
Length of the long thread, l = 18 cm = 18100=0.18 m=0.2 m.

Let I be the moment of inertia and ω be the angular speed.
Using the energy equation, we can write:
mgl(1-cos θ)+12Iω2=constant
mg0.20 1-cos θ+12Iω2=C   …1
Moment of inertia about the point of suspension A is given by,
I= 23mr2 + ml2  Substituting the value of l in the above equation,
we get : I=23m0.022+m 0.22 =23m0.0004+m0.04 =m0.00083+0.04 =m0.12083`

On substituting the value of I in equation (1) and differentiating it, we get:
ddtmg 0.2 1-cos θ+120.12083mω2=ddtc⇒mg0.2sinθdθdt+120.12083m×2ωdωdt=0 ⇒2sin θ=0.12083α because, g=10 m/s2⇒αθ=60.1208⇒ω2=49.66⇒ω=7.04
Thus, time period T will be:T=2πω=0.89 s

For a simple pendulum, time period (T) is given by,
T=2πlg
⇒T= 2π0.1810= 0.86 s

% change in the value of time period=0.89-0.860.89×100=0.3
∴ It is about 0.3% greater than the calculated value.


Answer: 52

It is given that:
Time period of oscillation, T = 2 s
Acceleration due to gravity, g = π2 ms−2
Let I be the moment of inertia of the circular wire having mass m and radius r.

(a) Time period of compound pendulum T is given by,
T=2Imgl=2Imgr   ∵l=r             …(1)
Moment of inertia about the point of suspension  is calculated as,
I = mr2 + mr2 = 2mr2
On substituting the value of moment of inertia I in equation (1), we get:
T=2π2mr2mgr=2π2rg⇒22π=2rg⇒2rg=1π2⇒r=g2π2     =0.5 m=50 cm

(b) From the energy equation, we have:
12Iω2-0=mgr 1-cos θ12Iω2-0=mgr 1-cos 2°

⇒122mr2·ω2=mgr1-cos 2°
∵I=2mr2⇒ω2=gr1-cos 2°On substituing the value of g and r in the above equation, we get:ω=0.11 rad/s               ⇒v=ω×2r=11 cms-1

(c) The acceleration is found to be centripetal at the extreme position.
Centripetal acceleration at the extreme position an is given by,
    an = ω2(2r) = (0.11) × 100 = 12 cm/s2
The direction of an is towards the point of suspension.

(d) The particle has zero centripetal acceleration at the extreme position.
However, the particle will still have acceleration due to the S.H.M.
Angular frequencyω is given by,
ω=2πT =2π2=3.14
∴ Angular acceleration a at the extreme position is given as,
α=ω2θα=ω22°=π2×2π180 =2π3180 1°=π180radian
Thus, tangential acceleration=α2r=2π3180×100
= 34 cm/s2


Answer: 53

It is given that:
Mass of disc = m
Radius of disc = r
The time period of torsional oscillations is T.
Moment of inertia of the disc at the centre, I=mr22

Time period of torsional pendulumT is given by,
T=2πIk
where I is the moment of inertia, and
 k is the torsional constant.

On substituting the value of moment of inertia in the expression for time period T, we have:
T=2πmr22k                On squaring both the sides, we get:T2=4π2mr22k=2π2mr2k⇒2π2mr2=kT2⇒k=2π2mr2T2


Exercise : Solution of Questions on page Number : 256


Answer: 54

It is given that the mass of both the balls is and they are connected to each other with the help of a light rod of length L.

Moment of inertia of the two-ball system I is given by,
I=2mL22=mL22


Torque τ, produced at any given position θ is given as:
τ = 
⇒ Work done during the displacement of system from 0 to θ0 will be,
W=∫0θ0kθ dθ=kθ022
On applying work-energy theorem, we get:
12Iω2-0=work done=kθ022∴ω2=kθ02I=kθ02mL2

From the free body diagram of the rod, we can write:
Force, T2=mω2L2+mg2  =mkθ02mL2×L2+m2g2 =k2θ04L2+m2g2


Answer: 55

It is given that a particle is subjected to two S.H.M.s of same time period in the same direction.

Amplitude of first motion, A1 = 3 cm
Amplitude of second motion, A2 = 4 cm

Let ϕ be the phase difference.

The resultant amplitude R is given by,
R=A12+A22+2A1A2 cos ϕ

(a) When ϕ = 0°
R=32+42+234 cos 0°  =7 cm

(b) When ϕ = 60°
R=32+42+234 cos 60° =37=6.1 cm

(c) When ϕ = 90°
R=32+42+234cos 90°  =25=5 cm


Answer: 56

It is given that three S.H.M.s of equal amplitudes and equal time periods are combined in the same direction.

Let Y1, Y2 and Y3 be the three vectors representing the motions, as shown in the figure given below.

According to the question:
Angle betweenY1 and Y2  = 60 °Angle between Y2 and Y3 = 60 °.


By using the vector method, we can find the resultant vector.
Resultant amplitude = Vector sum of the three vectors
= A + A cos 60° + A cos 60°
=A+A2+A2=2A


Answer: 57

Given are the equations of motion of a particle:
x1 = 2.0sin100πt
x2=2.0sin120πt+π3

The resultant displacement x will be,
x = x1 + x2
=2sin100πt+sin120πt+π3

(a) At t = 0.0125 s
x=2sin100π ×0.0125+sin120π ×0.0125+π3 =2sin 5π4+sin 3π2+π3 =2-0.707+-0.5 =2×-1.207=-2.41 cm

(b) At t = 0.025 s
x=2sin100 π×0.025+sin120π×0.025+π3 =2sin10π4+sin3π+π3 =21+-0.866 =2×0.134=0.27 cm


Answer: 58

Given:
Equation of motion along X axis, x = x0sinωt
Equation of motion along Y axis, s = s0sinωt
Angle between the two motions, θ = 45
Resultant motion (R) will be,
R=x2+s2+2xscos45° =x02sinωt+s02sinωt +2x0s0sin2ωt12 =x02+s02+2x0s01/2sinωt

Hence, the resultant amplitude is x02+s02+2x0s01/2.


error: