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HC Verma Class XII Physics Chapter 15 –Wave Motion and Wave on a String


Exercise : Solution of Questions on page Number : 321


Answer: 1

No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.


Answer: 2

It is a non-mechanical wave because this type of wave does not require a material medium to travel.


Answer: 3

Equation of the wave is
y=c1 sin (c2x+c3t)
When the variable of the equation is (c2x + c3t), then the wave must be moving in the negative x-axis with time t.


Answer: 4

Equation of the wave is given by
y=Asin(ωt−kx)
where


Answer: 5

When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored in the form of potential energy within it. After passing each other, both the pulses regain their original shape.


Answer: 6

No, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.


Answer: 7

Equation of the wave: y = A sin(kx − ωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.

The argument of the sine is a phase, so the smallest positive phase constant should be
sin(7.5π)=sin(3×2π+1.5π)
=sin(1.5π)
Therefore, the smallest positive phase constant is 1.5π.


Answer: 8

Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.


Exercise : Solution of Questions on page Number : 322


Answer: 1

(c) λ/2

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to λ/2.


Answer: 2

(c) λ/2

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to λ/2.


Answer: 3

(a) x=A sin ky-ωt

Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.


Answer: 4

(b) amplitude A/2, frequency ω/π

y=Asin2kx-ωt

cos2θ=1-2sin2θsin2θ=1-cos2θ2

y=A1-cos2kx-ωt2y=A21-cos2kx-ωt
Thus, we have:
Amplitude = A2
Frequency = 2ω2π=ωπ


Answer: 5

(d) Sound waves
There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.


Answer: 6

(a) ν

The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.


Answer: 7

(a) 1/2

Wave speed is given by
ν=Tµ
where
​T is the tension in the string
v is the speed of the wave
μ is the mass per unit length of the string
µ=ML=ρVL=ρALL
where
M is the mass of the string, which can be written as ρV
L is the length of the string
=ρπr2=ρπD24
∴ν=TρπD24=2DTρπ
where D is the diameter of the string.
Thus, v ∝ 1D
Since, rA = 2rB
vA∝12rA∝12×2rB (1)vB∝12rB (2)
From Equations (1) and (2) we get
vAvB=12


Answer: 8

(d) 1/2

TAB=TTCD=2T

where
​TAB is the tension in the string AB
​TCD is the tension in the string CD
The eelation between tension and the wave speed is given by
v=Tµv∝T
where
v is the wave speed of the transverse wave
μ is the mass per unit length of the string
v1v2=T2T=12


Answer: 9

(d) meaningless

Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.


Answer: 10

(c) λ'<λ

As v=fµ
A wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.


Answer: 11

(b) 2a

We know that the resultant of the amplitude is given by
Rnet=A12+A22+2A1A2 cosϕ
For the particular case, we can write
=a2+a2+2a2cosπ2=2a


Answer: 12

(d) the information is insufficient to find the relation between t1 and t2.

v=ηρ
But because the length of wires A and B is not known, the relation between A and B cannot be determined.


Answer: 13

(b) the velocity but not for the kinetic energy

The principle of superposition is valid only for vector quantities.
Velocity is a vector quantity, but kinetic energy is a scalar quantity.


Answer: 14

(d) The pulses will pass through each other without any change in their shapes.

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.


Answer: 15

(b) 2A2

We know resultant amplitude is given by
Anet=A12+A22+2A1A2cosϕ
For maximum resultant amplitude
Amax=A1+A2
For minimum resultant amplitude
Amin=A1-A2

So, the difference between Amax and Amin is
Amax-Amin=A1+A2-A1+A2=2A2


Answer: 16

(d) between 0 and 2A

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (Amax) and the minimum resultant amplitude (Amin).
Amax = A + A = 2A
Amin = A − A = 0


Answer: 17

(a) A

We know the resultant amplitude is given by
Rnet=A2+A2+2A2 cos 120º (ϕ=120°)=2A2-A2
∵ cos 120º=-12=A


Answer: 18

(a) inverse of its length

The relation between wave speed and the length of the string is given by
v=12lFµ
where
l is the length of the string
F is the tension
μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
v∝1l


Exercise : Solution of Questions on page Number : 323


Answer: 1

Given,

Speed of the wave pulse passing on a string in the negative x-direction = 40 cms−1
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
= 40 × 5
= 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.


Answer: 2

Given,
Equation of the wave travelling on a string stretched along the X-axis:
y=Ae xa+tT-2

(a) The dimensions of A (amplitude), T (time period) and a=λ2π, which will have the dimensions of the wavelength, are as follows:
A = M0L1T0T=M0L0T-1a=M0L1T0

(b) Wave speed, ν=λT=aT λ=a

(c) If y=f t+xν, then the wave travels in the negative direction; and if y=f t-xν, then the wave travels in the positive direction.
Thus, we have:
y=Ae xa+tT-2 =Ae-1Tt+xTa2 =Ae-1Tt+xV = Ae-ft+xV
Hence, the wave is travelling is the negative direction.

(d) Wave speed, v=at
Maximum pulse at t = T = aT×T=a Along the negative x-axis
Maximum pulse at t = 2T = aT×2T=2a Along the negative x-axis
Therefore, the wave is travelling in the negative x-direction.


Answer: 3

Given,
Wave pulse at t = 0

Wave speed = 10 cms−1
Using the formula s=v×t, we get:

At : t=1 s, s1=ν×t=10×1=10 cmt = 2 s,
s2=ν×t=10×2=20 cmt = 3 s,
s3=ν×t=10×3=30 cm


Answer: 19

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.


Answer: 20

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.


Answer: 21

(b) vibrate with a frequency of 208 Hz

According to the relation of the fundamental frequency of a string
ν=12lFμ
where
l is the length of the string
F is the tension
μ is the linear mass density

We know that ν1 = 416 Hz, l1 = l and l2 = 2l.
v1∝1l1v1l1=v2l2416l=v22lv2=208 Hz


Answer: 22

(d) 16 kg

According to the relation of the fundamental frequency of a string
ν=12lFμ
where l is the length of the string
F is the tension
μ is the linear mass density of the string
We know that ν1 = 416 Hz, l1 = l and l2 = 2l.
Also, m1 = 4 kg and m2 = ?
ν1=12l1m1gμ (1)
ν2=12l2m2gμ (2)
So, in order to maintain the same fundamental mode
ν1=ν2
squaring both sides of equations (1) and (2) and then equating
14l24gμ = 116l2m2gμ⇒m2=16 kg


Answer: 1

(c) may move on the X-axis
(d) may move on the Y-axis

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.


Answer: 2

(d) in the X–Y plane

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the X–Y plane.


Answer: 3

(d) be polarised

A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.


Answer: 4

(b) may be longitudinal
(d) may be transverse

Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.


Answer: 5

(a) must be longitudinal

Because particles in a gas are far apart, only longitudinal wave can travel through it.


Answer: 6

(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes.

A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.

yA→=Asinωt
yB→=Bsinωt+π


Answer: 7

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

y=0·001 mm sin50 s-1t+2·0 m-1x
Equating the above equation with the general equation, we get:
y=Asinωt-kxω=2πT=2πvk=2πλ
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
A=0.001 mmNow,50=2πν⇒ν=25π Hz


Answer: 8

(a) must be an integral multiple of λ/4

A standing wave is produced on a string clamped at one end and free at the other.
Its fundamental frequency is given by
ν=n+12v2L⇒v=νλ⇒ν=n+12νλ2L⇒L=2n+14λ⇒L=λ4,3λ4,…


Answer: 9

(b) The energy of any small part of a string remains constant in a standing wave.

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.


Answer: 10

(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

Thus, particles in alternate antinodes vibrate in the same phase.


Exercise : Solution of Questions on page Number : 324


Answer: 4

Given,
Pulse travelling on a string, y=a3x-νt2+a2
a=5 mm=0.5 cmWave speed, ν = 20 cm/s
So, at t=0 s, y=a3x2+a2.
Similarly, at t = 1 s,
y=a3x-ν2+a2And, At t=2 s,y=a3x-2ν2+a2

To sketch the shape of the string, we have to plot a graph between y and x at different values of t.


Answer: 5

Given,
Equation of the wave travelling in the positive x-direction at x = 0:
ft=AsintT
Here,
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by
y=AsintT-xνT


Answer: 6

The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
Now,
(a) M0L1T0=A⇒A=LAnd, a=M0L1T0⇒a=L(b) Wave speed=ν
∴ Time period, T=aνHere, a=Wave length=λ
The general equation of wave is represented byy=Asinxa-tav =Asinx-νta


Answer: 7

Given,
Wave velocity = ν
Shape of the string at t=t0 = gx, t0=A sin x/a …(i)
For a wave travelling in the positive x-direction, the general equation is given by
y=A sin xa-tT
Putting t = − t and comparing with equation (i),
we get:
gx,0=Asinxa+t0T⇒gx,t=Asinxa+t0T-tTNow,T=aν Here, a=Wave lengthν=Velocity of the wave
Thus, we have : y=Asin xa+t0aν-taν

⇒y=Asin x+ν t0-ta


Answer: 8

Given,
Equation of the wave, y=0.10 mm sin31.4 m-1x+314 s-1 t
The general equation is y=Asin2πxλ+ωt.
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b) 2πλ=31.4 m-1
⇒λ=2π31.4=0.2 m= 20 cmAnd, ω=314 s-1⇒2πf=314⇒f=3142π =3142×3.14 =50 s-1=50 Hz

Wave speed:
ν=λf=20×50 = 1000 cm/s

(c) Maximum displacement, A = 0.10 mm
Maximum velocity=aω=0.1×10-1×314 =3.14 cm/s


Answer: 9

A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s

(a) General wave equation along the x-axis:
y=Asinkx-ωt
∴k=2πλ=2π2=π cm-1T=λν=22000 =11000=10-3 sω=2πT=2π×103 s-1
Wave equation:
y=0.2 cm sinπ cm-1 x-2π×10-3 s-1

(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.
y=0.2 cm sin2π=0
∴ν=Aωcosπx =0.2×2000π×cos2π =400π =400π cm/s=4π m/s

If the wave equation is written in a different fashion, then also we will get the same values for these quantities.


Answer: 10

The wave equation is represented by
y=1·0 mm sin πx2·0 cm-t0·01 s
Let:
Time period = T
Wavelength = λ

a T=2×0.01=0.02 s=20 ms λ=2×2=4 cm

(b) Equation for the velocity of the particle:
ν=dydt=ddtsin 2πx4-t0.02
=-0.50 cos 2πx4-t0.02×10.02⇒ν=-0.50 cos 2π x4-t0.02At x=1 and t=0.01 s, ν=-0.50 cos 2π 14-12=0.

(c) (i) Speed of the particle:
At x=3 cm and t=0.01 s, ν=-0.50cos2π34-12=0.

(ii) At x=5 cm and t=0.01 s,
ν=0 iii At x=7 cm and t=0.1 s, ν=0.iv At x=1 cm and t=0.011 s,ν=50 cos 2π14-0.0110.02 =-50 cos 3π5=-9.7 cm/s
(By changing the value of t, the other two can be calculated.)
Answer: 11

Time taken to reach from the mean position to the extreme position, T4 = 5 ms
Time period (T) of the wave:
T=4×5 ms =20×10-3=2×10-2 s
Wavelength (λ) = 2×Distance between two mean positions
=2×2 cm=4 cm
Frequency, f=1T =12×10-2 =50 Hz
Wave speed, v= λf =4×10-2×50 =200×10-2 =2 m/s


Answer: 12

Given:
Wave speed, ν=20 cm/s
From the graph, we can infer:
(a) Amplitude, A = 1 mm

(b) Wavelength, λ = 4 cm

(c) Wave number, k=2πλ
=2×3.144=1.57 cm-2

(d) Time period, T=λν
Frequency, f=1T=vλ⇒f=204=5 Hz


Answer: 13

Given,
Wave speed (v) = 10 ms−1
Time period (T) = 20 ms =20×10-3=2×10-2 s

(a) Wavelength of the wave:
λ=νt=10×2×10-2 = 0.02 m=20 cm

(b) Displacement of the particle at a certain instant:
y=asinωt-kx⇒1.5=asinωt-kx
Phase difference of the particle at a distance x = 10 cm :
ϕ=2πxλ=2π×1020=π
The displacement is given by y’=asinωt-kx+π =asinωt-kx=1.5 mm∴Displacement=1.5 mm


Answer: 14

Given,
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:
Mass per unit length=564 gm/cmTension, T=8 N =8×105 dynSpeed, v=Tm =8×105×645 =3200 cm/s=32 m/s


Answer: 15

Given,

Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm−1
Applied tension = 16 N = 16×105 dyn
Velocity of the wave:
ν=Tm =16×1050.4 =2000 cm/s
∴ Time taken to reach the other end =202000=0.01 s
Time taken to see the pulse again in the original position=0.01×2=0.02 s

(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.


Answer: 16

Given,
Linear mass density of the string = 0.5 gcm−1
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms−1

The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.

Phase change=0
(a) Total distance, S=20+20=40 cmWave speed, ν=20 m/s
Time taken to regain shape:
Time=Sν=4020=2 s

(b) The wave regain its shape after covering a period distance=2×30=60 cm
∴ Time period=6020=3 s

(c) Frequency, n=1Time period=13 s-1
We know:
n=12lTm
Here, T is the tension in the string.
Now,

m=Mass per unit length =0.5 gm/cm⇒13=12×30 T0.5⇒ T=400×0.5 =200 dyn =2×10-3 N


Answer: 17

Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v1.
Thus, we have:
​ν1=Tm1
The mass per unit length can be given as
m1=ρ1a1I1I1=ρ1a1⇒ν1=Tρ1a1 …(1)
Let the velocity of the first string be v2.
Thus, we have:
ν2=Tm2⇒ ν2=Tρ2a2 …(2)
Given,
ν1=2ν2⇒Ta1ρ1=2Ta2ρ2⇒Ta1ρ1=4 Ta2ρ2⇒ρ1ρ2=14⇒ρ1:ρ2 = 1:4


Exercise : Solution of Questions on page Number : 325


Answer: 18

Given,
Wave equation, y=0·02 msin1·0 m-1x+30 s-1t

Let: Mass per unit length, m=1.2×10-4 kg/mFrom the wave equation, we have:k=1 m-1=2πλAnd,ω=30 s-1=2πf

Velocity of the wave in the stretched string is given by

ν=λf=ωk=301⇒v=30 m/sWe know:v=Tm⇒30 =T1.2×10-4 ⇒T=108×10-3=0.108 N

So, the tension in the string is 0.108 N.


Answer: 19

Given,
Amplitude of the wave = 1 cm
Frequency of the wave, f=2002=100 Hz
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by
v=Tm
Thus, we have:
v=900.1=30 m/sNow,Wavelength, λ=vf=30100=0.3 m⇒λ=30 cm

(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
y=1 cmcos2πt0.01 s-x30 cm …(1)

(c) Using cos-θ=cosθ in equation (1), we get:
y=1cos2πx30-t0.01
Velocity, v=dydt⇒v=2π0.01sin2πx30-t0.01And,Acceleration, a=dνdt⇒a=4π20.012cos2πx30-t0.01When x=50 cm, t=10 ms=10×10-3 s.

Now,

v=2π0.01sin2π53-0.010.01 =2π0.01sin2π×23 =-2π0.01sin4π3 =- 200πsinπ3 =-200π×32 =-544 cm/s =-5.4 m/s
In magnitude, v = 5.4 m/s.
Similarly,
a=4π20.012cos2π53-1 =4π2×104×12 ≈2×105 cm/s2 or 2 km/s2


Answer: 20

Given,
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length=1040=14 gm/cm
Spring constant, k = 160 N/m

Deflection, x=1 cm =0.01 mTension, T= kx=160 ×0.01⇒T=1.6 N=16×104 dynNow, v=Tm=16×10414⇒v=8×102 cm/s=800 cm/s

∴ Time taken by the pulse to reach the spring, t=40800=120=0.05 s


Answer: 21

Given,
Mass of each block = m1=m2=3.2 kg
Linear mass density of wire AB = 10 gm−1 = 0.01 kgm−1
Linear mass density of wire CD = 8 gm−1 = 0.008 kgm−1
For string CD, velocity is defined as v=Tm.
Here, T is the tension and m is the mass per unit length.
For string CD,
T=3.2×g
Thus, we have:
v =3.2×100.008 =32×1038 =2×1010 =20×3.14≈63 ms

For string AB,
T=2×3.2g=64 NThus, we have:v=Tm =640.01=6400 =80 m/s


Answer: 22

Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:
m=4.5×10-32.25 =2×10-3 kg/mT=2g=20 NWave speed, ν=Tm =202×10-3 = 104 = 102 m/s=100 m/s

Time taken by the disturbance to reach the pulley:
t=sν =2100=0.02 s


Answer: 23

Given,
Mass of the block = 4.0 kg
Linear mass density, m=19.2×10-3 kg/m
From the free body diagram,
T-4g-4a=0⇒T=4a+g =42+10=48 N

Wave speed, ν=Tm =4819.2×103 =2.5×10-3=50 m/s


Answer: 24

Given,
Speed of the transverse pulse when the car is at rest, v1 = 60 cm s−1
Speed of the transverse pulse when the car accelerates, v2= 62 cm s−1
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,
Wave speed, ν=Tm=MgmWhen car is at rest:Tension in the string, T=Mg⇒v1=Mgm⇒Mgm=602 …(i)
When car is having acceleration:
Tension, T=Ma2+Mg2
Again, ν2=Tm⇒62=Ma2+Mg21/4m1/2
⇒ Ma2+Mg2m=622 …(ii)
From equations (i) and (ii), we get:
Mgm×mMa2+Mg2=60622⇒ga2+g2=0.936⇒g2a2+g2=0.876⇒a2+100 0.876=100⇒ a2=12.40.867=14.15⇒a=3.76 m/s2Therefore, acceleration of the car is 3.76 m/s2.


Answer: 25

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity

Consider one half of the string, as shown in the figure.

The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part dθ at angle θ.
So,
Length of the element=Rdθ, mass =mRdθ
Centrifugal force experienced by the element=mRdθ ω2R
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
dF=2mR2dθω2 sinθ
Total force, F=∫0π/22mR2ω2sinθdθ =2mR2ω2 -cos θ =2mR2ω2And, 2T=2mR2ω2⇒T=mR2ω2
Velocity of the transverse vibration is given as
V’=TmV’=mR2ω2m=ωR
Linear velocity of the string, V = ωR
∴ Speed of the disturbance, V’ = V


Answer: 26

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
v=Tm=mgxm⇒v=gx

(b) Let the time taken be dt for the small displacement dx.
Thus, we have:

dt=dxv=dxgx
∴Total time, T=∫0Ldxgx=4Lg

(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.

Now,
t=∫0ydxgx =4yg

∴ Distance travelled by the particle in this time, S = L-y
Using the equation of motion, we get:
S=ut+12 gt2⇒L-y=12 g×4yg2⇒L-y=2y⇒3y=L⇒y=L3
Thus, the particle will meet the pulse at a distance L3 from the lower end.


Answer: 27

Given,
Linear density of each of two long strings A and B, m = 1.2×10-2 kg/m
String A is stretched by tension Ta= 4.8 N.
String B is stretched by tension Tb= 7.5 N.
Let va and vb be the speeds of the waves in strings A and B.
Now,
va=Tam⇒va=4.81.2×10-2=20 m/svb=Tbm⇒vb=7.51.2×10-2=25 m/st1=0 in string At2=0+20 ms=20×10-3=0.02 s
Distance travelled by the wave in 0.02 s in string A:
s=20×0.02=0.4 m
Relative speed between the wave in string A and the wave in string B, v’=25-20=5 m/s
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m
t’=sv’=0.45=0.08 s


Answer: 28

Given,
Amplitude of the transverse wave, r = 0.5 mm =0.5×10-3 m
Frequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:
ν=Tm⇒ν2=Tm⇒m=Tν2=1001002 =0.01 kg/mAverage power of the source:Pavg=2π2mνr2f2 =2 3.142 0.01×100×0.5×10-32×100 =2×9.86×0.25×10-6×104 =19.7×0.0025=0.049 W =49×10-3 W=49 mW


Answer: 29

Given,
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10−3 m
Linear mass density, m = 6 gm−3
Applied tension, T = 60 N
Now,
Let the velocity of the wave be v.
Thus, we have:
v=Tm=606×10-3=102=100 m/s

(a) Average power is given as
Paverage=2π2mνA2f2 =2×3.142×6×10-3×100×10-3×2002 =473×10-3=0.47 W

(b) Length of the string = 2 m
Time required to cover this distance:
t=2100=0.02 sEnergy=Power×t =0.47×0.02 =9.4×10-3 J=9.4 mJ


Answer: 30

Given,
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm−1
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be λ.

(a) The speed of the transverse wave is given by
ν=Tm
⇒v=490.01=70 m/s
Also, ν=fλ∴ λ=fv=70440=16 cm

(b) Maximum speed (vmax) and maximum acceleration (amax):
We have: y=A sin ωt-kx
∴ ν=dydt=Aω cos ωt-kxNow,νmax=dydt=Aω =0.50×10-3×2π×440 =1.3816 m/s.And, a=d2ydt2⇒a=-Aω2 sin ωt-kxamax=-Aω2 =0.50×10-3×4π2 4402 =3.8 km/s2

(c) Average rate (p) is given by
p=2π2νA2f2 =2×10×0.01×70×0.5×10-32×4402 =0.67 W


Answer: 31

Given,
Phase difference between the two waves travelling in the same direction, ϕ=90°=π2.
Frequency f and wavelength λ are the same. Therefore, ω will be the same.
Let the wave equations of two waves be:
y1=rsinωt
y2=rsinωt+π2
Here, r is the amplitude.
From the principle of superposition, we get:
y=y1+y2 =rsinωt+rsinωt+π2 =r sinωt+sinωt+π2 =r2sinωt+ωt+π22cosωt-ωt-π22 =2rsinωt+π4cos-π4 =2rsinωt+π4
∴ Resultant amplitude, r’=2r=42 mm


Answer: 32

Given,

Speed of the wave pulse travelling in the opposite direction, v = 50 cm s−1 = 500 mm s−1
Distances travelled by the pulses:
Using s = vt, we get:
In t=4 ms=4×10-3 s,s=νt=500×4×10-3=2 mm.In t=6 ms=6×10-3 s,s=500×6×10-3=3 mm.In t=8 ms=8×10-3 s,s=νt=500×8×10-3= 4 mm.In t=12 ms=12×10-3 s,s=500×12×10-3=6 mm.

The shapes of the string at different times are shown in the figure.


Exercise : Solution of Questions on page Number : 326


Answer: 33

Given:
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm =2×10-2 m
Wave speed, ν =f×λ=100×2×10-2 m/s =2 m/s

(a) First wave will travel the distance in 0.015 s.
⇒x=0.015×2 =0.03 m
This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:
ϕ=2πxλ =2π2×10-2×0.03=3π

(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:
⇒ϕ=2πxλ =2π2×10-2×0.04 =4π

(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:
y1=r sin ωtAnd, y2=r sin ωt+ϕBy using the principle of superposition:y=y1+y2 =rsinωt+ωt+ϕ =2rsinωt+ϕ2 cosϕ2
∴ Resultant amplitude =2r cosϕ2
So, when ϕ=3x:⇒r=2×10-3 mRresultant=2×2×10-3 cos 3π2 =0
Again, when ϕ=4π:Rresultant=2×2×10-3 cos 4π2 =4×10-3×1 =4 mm


Answer: 34

Length of a stretched string (L) = 1 m
Wave speed (v) = 60 m/s
Fundamental frequency (f0) of vibration is given as follows:
f0=v2L
=602×1=30 s-1=30 Hz


Answer: 35

Given:
Length of the wire (L)= 2.00 m
Fundamental frequency of the vibration (f0) = 100 Hz
Applied tension (T) = 160 N

Fundamental frequency, f0=12LTm⇒10=14160m⇒m=1×10-3 kg/m⇒m=1 g/m
So, the linear mass density of the wire is 1 g/m.


Answer: 36

Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m) =480 g/cm =0.005 kg/m

Wave speed, ν=Tm =500.005=100 m/s

Fundamental frequency, fo=12LTm =12×0.8×500.005 =1002×0.8=62.5 HzFirst harmonic=62.5 HzIf f4= frequency of the fourth harmonic:⇒f4=4f0=62.5×4⇒f4=250 HzWavelength of the fourth harmonic, λ4=νf4=100250⇒λ4=0.4 m=40 cm


Answer: 37

Given:
Length of the piano wire (L)= 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (fo) = 261.63 Hz

Linear mass density, m=690 gm/cm =6×10-390×10-2 kg/m =6900 kg/m

Fundamental frequency, fo=12LTm
⇒ 261.63=12×0.09 T×9006⇒0.18×261.63=150 T⇒150 T=261.63×0.182⇒T=261.63×0.182150 =1478.52 N≈1480 N

Hence, the tension in the piano wire is 1480 N.


Answer: 38

Given:
Length of the sonometer wire (L) = 1.50 m
Let the first harmonic be f0 and the second harmonic be f1.
According to the question, f1 =256 Hz
1st harmonic for fundamental frequency, f0=f12=2562=128 Hz

When the fundamental wave is produced, we have:
λ2=1.5 m⇒ λ=3 m
Wave speed, v=f0λ⇒v=128×3=384 m/s


Answer: 39

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm

Mass per unit length, m=121.5 g/m =8×10-3 kg/m

Tension in the wire, T=9×g =90 N

Fundamental frequency is given by:

f0=12L Tm
For second harmonic (when two loops are produced):
f1=2f0=11.5 908×10-3 =106.061.5 =70.7 Hz≈70 Hz


Answer: 40

Given:
Length of the stretched string (L) = 1.00 m
Mass of the string =40 g
String is attached to the tuning fork that vibrates at the frequency (f) = 128 Hz
Linear mass density (m) =40×10-3 kg/m
No. of loops formed, (n) = 4
L=nλ2⇒λ=2Ln=2×14⇒λ=0.5 mWave speed (v)=fλ=128×0.5⇒v=64 m/s

We know: v=Tm⇒T=ν2m =642×40×10-3 =163.84≈164 N
Hence, the tension in the string if it is to vibrate in four loops is 164 N.


Answer: 41

Given:
Wire makes a resonant frequency of 240 Hz and 320 Hz when its both ends are fixed.
Therefore, fundamental frequency (f0) of the wire must be the factor of 240 Hz and 320 Hz.
(a) Maximum value of fundamental frequency, f0 = 80 Hz

(b) Wave speed (v) = 40 m/s
And if λ is the wavelength:
λ2=L

∴ v=λ×f0⇒v=2×L×f0⇒L=402×80⇒L=14 m=0.25 m


Answer: 42

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be ‘n’ loops and λ be the wavelength.
∴ λ = 2×Separation between the consecutive nodes

λ2=2×1.6=3.2 cm

Length of the wire is L.

In the first case:
L=nλ12
In the second case:
L=n+1λ22⇒nλ12=n+1 λ22⇒n×4=n+13.2⇒4n-3.2n=3.2⇒0.8 n=3.2⇒n=4
∴ Length of the string, L=nλ12=4×42=8 cm


Answer: 43

Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s

Wavelength, λ=vf=220660=13 m
(a) No. of loops, n = 3
∴ L=n2λ
⇒L=32×13⇒L=12 m=50 cm

(b) Equation of resultant stationary wave can be given by:
y=2Acos2πxλsin2πvLλ⇒y=0.5 cos2πx13sin2π×220×t13⇒y=0.5 cm cos6πx m-1 sin1320πt s-1


Answer: 44

Given:
Length of the guitar wire (L1) = 30.0 cm = 0.30 m
Frequency, when no finger is placed on it, (f1) =196 Hz
And (f2) =220 Hz, (f3) = 247 Hz, (f4) = 262 Hz and (f5) = 294 Hz
The velocity is constant for a medium.
We have:
f∝1L
⇒ f1f2=L2L1⇒196220=L20.3⇒L2=196×0.3220=0.267 m⇒L2=26.7 cm

Again, f3=247 Hz
⇒ f3f1=L1L3⇒247196=0.3L3⇒L3=196×0.3247=0.238 m⇒L3=23.8 cmSimilarly, L4=196×0.3262=0.224 m⇒L4=22.4 cmAnd, L5=20 cm


Answer: 45

Given:
Fundamental frequency (f0) of the steel wire = 200 Hz
Let the highest harmonic audible to the person be n.

Frequency of the highest harmonic, f’ = 14000 Hz
∴ f’= nf0 …(1)
f’f0=14000200⇒nf0f0=70⇒ n=70
Thus, the highest harmonic audible to man is the 70th harmonic.


Answer: 46

Given:
Let the three resonant frequencies of a string be
f1=90 Hzf2=150 Hzf3=210 Hz

(a) So, the highest possible fundamental frequency of the string is f=30 Hz, because f1, f2 and f3 are the integral multiples of 30 Hz.

(b) So, these frequencies can be written as follows:
f1=3ff2=5ff3=7f

Hence, f1, f2, and f3 are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f …
∴ 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic

(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:
f1=32×L v⇒90=32×80×v⇒v=90×2×803 =30×2×80 =4800 cm/s⇒v=48 m/s


Answer: 47

Given:
The tensions in the two wires are in the ratio of 2:1.
⇒T1T2=2
Ratio of the radii is 3:1.
⇒r1r2=3=D1D2
Density in the ratios of 1:2.
⇒ρ1ρ2=12
Let the length of the wire be L.
Frequency, f=1LDTπρ
⇒f1=1LD1T1πρ1⇒f2=1LD2T2πρ2
∴ f1f2=LD2LD1T1T2πρ2πρ1⇒f1f2=1321×21⇒ f1:f2=2:3


Answer: 48

Given:
Length of the rod (L) = 40 cm = 0.40 m
Mass of the rod (m) = 1.2 kg
Let the mass of 4.8 kg be placed at x distance from the left.
As per the question, frequency on the left side = f0
Frequency on the right side = 2f0
Let tension be T1 and T2 on the left and the right side, respectively.

∴ 12LT1m=22LT2m⇒T1T2=2⇒T1T2=4 … (1)
From the free body diagram:

T1+T2=48+12=60 N⇒ 4T2+T2=5T2=60 N using equation 1∴T2=12 Nand T1=48 N

Now, taking moment about point A:


T2×0.4=48x+12×0.2⇒4.8=48x-2.4⇒4.8x=2.4⇒x=2.44.8=120 m=5 cm

Therefore, the mass should be placed at a distance of 5 cm from the left end.


Answer: 49

Given:
Length of the aluminium wire (La)= 60 cm = 0.60 m
Length of the steel wire (Ls)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (Aa) = 1.0 mm2
Area of cross-section of the steel wire (As) = 3.0 mm2
Density of aluminium (ρa) = 2⋅6 g cm−3
Density of steel (ρs) = 7⋅8 g cm−3

Mass per unit length of the steel, ms =ρs×As =7.8×10-2 gm/cm =7.8×10-3 kg/mMass per unit length of the aluminium, mA=ρAAA =2.6×10-2×3 gm/cm =7.8×10-2 gm/cm

= 7.8×10-3 kg/m

A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.
⇒v=Tm =407.8×10-3- =4×1047.8⇒v=71.6 m/s
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm
⇒ Wavelength, λ=2×20=40 cmOr λ=0.4 m∴Frequency, f=vλ=71.60.4=180 Hz


Answer: 50

Given:
Length of the string = L
Velocity of wave is given as:
V=Tm
In fundamental mode, frequency = ν
⇒ν=12LTm
(a) Wavelength, λ=VelocityFrequency
⇒λ=Tm12LTm=2LWave number, K=2πλ=2π2L=πL

(b) Equation of the stationary wave:
y=Acos2πxλsin2πVtλ =Acos2πx2Lsin2πVt2L ∵ ν=V2L =AcosπxLsin2πνt


Exercise : Solution of Questions on page Number : 327


Answer: 51

Given:
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm

(a) Wavelength and frequency of the string when it is vibrating in its 1st overtone (n = 2):
L=nλ2
⇒λ=L=2 m⇒f=νλ=2002=100 Hz
(b) The stationary wave equation is given by:

y=2A cos2πxλsin2πvtλ =0.5cos2πx2sin2π×200 t2 =0.5 cmcosπm-1 xsin200πs-1 t


Answer: 52

Given:
The stationary wave equation of a string vibrating in its third harmonic is given by
y = (0.4 cm) sin [(0.314 cm−1) x]cos [(.600 πs−1) t]
By comparing with standard equation,
y = A sin (kx) cos (wt)

(a) From the above equation, we can infer the following:
ω=600 π
⇒2πf=600 π⇒f=300 Hz
Wavelength, λ=2π0.314=2×3.140.314
⇒λ=20 cm

(b) Therefore, the nodes are located at 0cm, 10 cm, 20 cm, 30 cm.

(c) Length of the string, l = nλ2

⇒l=3λ2=3×202=30 cm

(d) y=0.4 sin 0.314 x cos 600 πt
=0.4sinπ10 xcos600πt

λ and ν are the wavelength and velocity of the waves that interfere to give this vibration.

λ=20 cmν=ωk=600 ππ10=6000 cm/s⇒ν=60 m/s


Answer: 53

Given:
Equation of the standing wave:
y=0.4 cm sin 0.314 cm-1 xcos 600 πs-1 t⇒k=0.314=π10Also, k=2πλ⇒λ=20 cm

We know:
L=nλ2
For the smallest length, putting n = 1:
⇒L=λ2=20 cm2=10 cm
Therefore, the required length of the string is 10 cm.


Answer: 54

Given:
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:
m=0.00320.4=8×10-3 kg/mChange in length, ∆L=40.05-40=0.05 cm =0.05×10-2 mStrain=∆LL=0.05×10-20.4 =0.125×10-2f0=12LTm =12×0.4005 T8×10-3

⇒220×220=10.8012×T×1038⇒T×1000=220×220×0.641×0.8⇒T=248.19 NStress=TensionArea=248.191 mm2=248.1910-6⇒Stress=248.19×106Young’s modulus, Y=stressstrain =248.19×1060.125×10-2⇒Y=19852×108 =1.985×1011 N/m2
Hence, the required Young’s modulus of the wire is 1.985×1011 N/m2.


Answer: 55

Density of the block = ρ
Volume of block = V
∴ Weight of the block is, W = ρVg
∴ Tension in the string, T = W

The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be f10.

f10=102LTm =102L ρ Vgm
Here, m is the mass per unit length of the string and L is the length of the string.

When the block is immersed in water (density = ρ ), let the eleventh harmonic be f11.

f11=112LT’m =112Lρ-ρw Vgm
The frequency (f) of the tuning fork is same.
∴ f10=f11⇒102LρVgm=112Lρ-ρω Vgm⇒100121=ρ-1ρ because ρω=1 gm/cc⇒100 ρ=121 ρ-121⇒ρ=12121=5.8 gm/cc =5.8×103 kg/m3
Therefore, the required density is 5.8×103 kg/m3.


Answer: 56

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, m =0.082=0.04 kg/m
Tension, T=256 NWave velocity, v=Tm⇒v=2560.04=1602⇒v=80 m/s

For fundamental frequency:
L=λ4⇒λ=4L=4×2=8 m⇒f=vλ=808=10 Hz

(a) The frequency overtones are given below:
1st overtone=3f=30 Hz2nd overtone=5f=50 Hz

(b) λ=4l=4×2=8 m
∴ λ1=vf1=8030=2.67 m λ2=vf2=8050=1.6 m

Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.


Answer: 57

Let T be the tension in the string and m be the mass per unit length of the heavy string.
In the first part of the question, the heavy string is fixed at only one end.
So, the lowest frequency is given by:
f0=14LTm …1
When the movable support is pushed by 10 cm to the right, the joint is placed on the pulley and the heavy string becomes fixed at both the ends (keeping T and m same).
Now, the lowest frequency is given by:
f0’=12LTm …2
Dividing equation (2) by equation (1), we get:


 

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