IISER Physics - Wave Motion
Exam Duration: 45 Mins Total Questions : 30
If L1 and L1 are the lengths of first and second harmonics of a resonance tube, then the wavelength of the wave is
- (a)
2(L1+L2)
- (b)
2(L2-L1)
- (c)
\(2(L_2-{L_1\over2})\)
- (d)
\(2({L_1\over2}+L_2)\)
If L1 and L2 are the first and second resonances, then we have
L1=\(\lambda\over4\) and L2=\(3\lambda\over4\)
L2-L1=\({\lambda\over2}\Rightarrow \lambda=2(L_2-L_1)\)
A policeman on duty detects a drop of 15% in the pitch of horn of motor car as it crosses him.If velocity of sound is 330m/s.Calculate speed of car.
- (a)
28.2m/s
- (b)
26.7m/s
- (c)
29.3m/s
- (d)
22.3m/s
Before crossing policeman, source is moving towards listener.
\(\therefore \ v^{ ' }=\frac { vv }{ v-{ v }_{ s } } \) ....................................(i)
After crossing the policeman, source is moving away from listener.
\(\therefore \ v^{ '' }=\frac { vv }{ v+{ v }_{ s } } \)
Dividing Eq. (ii) by Eq. (i), we get
\(\frac { v^{ '' } }{ v^{ ' } } =\frac { v-{ v }_{ s } }{ v+{ v }_{ s } } \)
Drop of 15% means
\(\frac { 85 }{ 100 } =\frac { { v }^{ '' } }{ { v }^{ ' } } \Rightarrow \frac { 85 }{ 100 } =\frac { 330-{ v }_{ s } }{ 330+{ v }_{ s } } \)
vs = 26.7 m/s
If the intensity of sound increases by a factor of 105, then increases in the intensity level is
- (a)
5dB
- (b)
10dB
- (c)
25dB
- (d)
50dB
Given, \(\frac { { l }_{ 2 } }{ { l }_{ 1 } } ={ 10 }^{ 5 }\Rightarrow \frac { { l }_{ 2 }/{ l }_{ 0 } }{ { l }_{ 1 }/{ l }_{ 0 } } ={ 10 }^{ 5 }\)
where, l0 s reference intensity
\(\Rightarrow log_{ 10 }\left( \frac { { l }_{ 2 } }{ { l }_{ 0 } } \right) -log_{ 10 }\left( \frac { { l }_{ 1 } }{ { l }_{ 0 } } \right) =5\)
\(\Rightarrow 10log_{ 10 }\left( \frac { { l }_{ 2 } }{ { l }_{ 0 } } \right) -10log_{ 10 }\left( \frac { { l }_{ 1 } }{ { l }_{ 0 } } \right) =50\)
\(\Rightarrow 10\left[ log_{ 10 }\left( \frac { { l }_{ 2 } }{ { l }_{ 0 } } \right) -log_{ 10 }\left( \frac { { l }_{ 1 } }{ { l }_{ 0 } } \right) \right] =50\)
\(\Rightarrow \ dB_{ 2 }-dB_{ 1 }=50 \ dB\)
A cylindrical tube, open at both ends, has a fundamental frequency, 'f' in air.The tube is dipped vertically in water, so that half of it is in water.The fundamental frequency of the air-column is now,
- (a)
f
- (b)
\(f\over2\)
- (c)
\(3f\over4\)
- (d)
2f
Initially for open organ pipe, fundamental frequency,
V0= \(v\over2l\)=f
But when it is half dipped in water, then it becomes closed organ pipe of length\(l\over2\)In his case, fundamental frequency,
\(v_c={v\over4l'}={v\over4{l\over2}}={v\over2l}=f\)
When two tuning forks 9fork 1 and fork 2) are sounded simultaneously, 4beat/s are heard.Now some tape is attached on the prong of the fork2.When the tuning forks are sounded again, 6 beat/s are heard.If the frequency of fork 1 is 200Hz, then what was the original frequency of fork 2?
- (a)
200Hz
- (b)
202Hz
- (c)
196Hz
- (d)
204Hz
The frequency of fork 2 = 200± 4 = 204 Hz or 196Hz
Since, on attaching the tape on the prong of fork 2, its frequency decreases, but now number of beats per second is 6 i.e., the frequency difference now increases. It is possible only when before attaching the tape, the frequency of fork 2 is less than the frequency of tuning fork 1. Hence, the frequency of fork 2 is 196 Hz.
A light pointer fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. Eight complete oscillations are counted when the plate falls through 10 em. What is the frequency of the tuning fork?
- (a)
112 Hz
- (b)
56 Hz
- (c)
8/7 Hz
- (d)
7/8 Hz
The time taken by the plate falling through a distance y is given by:
\(t=\sqrt { (2y/g) } =\sqrt { (2\times 10/980) } =(1/7)sec.\)
The number of oscillations completed in 117see is 8.
∴ Frequency = number of oscillations completed in one see
\(=\frac{8}{1/7}\)=56 Hz.
Sound waves in air are always longitudinal, because:
- (a)
density of air is very small
- (b)
air is a mixture of several gases
- (c)
air does not have a modulus of rigidity
- (d)
of the inherent characteristics of sound waves in air
We know that longitudinal waves travel in an elastic medium in the form of compressions and rarefactions which changes its volume. elasticity and pressure. Since, air is completely an elastic medium, i. e., it does not have a modulus of rigidity, therefore sound waves in air are always longitudinal.
The amplitude of a wave disturbance propagating in the positive Y-direction is given by:
\(y=\frac { 1 }{ 1+{ x }^{ 2 } } \)at t=0 and y=\(y=\frac { 1 }{ [1+{ (x }-1)^{ 2 }] } \)at t=2sec.Where, x and yare in m. If the shape of the wave disturbance does not change during the propagation, what is the velocity of the wave?
- (a)
1 m/sec
- (b)
1.5 m/sec
- (c)
0.5 m/sec
- (d)
2 m/sec
At t=0, \(y=\frac { 1 }{ 1+{ x }^{ 2 } } \) or\(x=\sqrt { \frac { 1-y }{ y } } ={ x }_{ 1 }\)
\(At\quad t=2sec,\quad y=\frac { 1 }{ \left[ 1+({ x-1) }^{ 2 } \right] } or\quad x=1+\sqrt { \frac { 1-y }{ y } } ={ x }_{ 1 }\)
\(\therefore \quad v=\frac { { x }_{ 2 }-{ x }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } =\frac { 1+\sqrt { \frac { 1-y }{ y } } -\sqrt { \frac { 1-y }{ y } } }{ 2-0 } =\frac { 1 }{ 2 } =0.5m/s\)
The equation y = a sin 2π\(\left( \frac { t }{ T } -\frac { x }{ \lambda } \right) \) of a simple harmonic wave gives us:
- (a)
the displacement of all particles of the medium at a particular instant of time only
- (b)
the displacement of a single particle at any time
- (c)
the displacement of all the particles of the medium at a particular instant of time as well as the displacement of a single particle at any time
- (d)
the behaviour of the medium as a whole
The kinetic energy of a particle executing SHM is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm and mass of the particle is 5.12 kg, the time period of its oscillation is:
- (a)
(π/5)sec
- (b)
2π sec
- (c)
20π sec
- (d)
5π sec
A siren can be made from a rotating flat disc, which has regularly spaced holes punched through it along a circle concentric with the axis of rotation. An air nozzle is directed against the disc. Each time a hole passes the nozzle, a puff of air is released to generate a wave pulse. Sound of what frequency will be produced by disc containing 72 holes and rotating at 1800 rev./min.?
- (a)
72x 1800Hz
- (b)
72x 1800x 60Hz
- (c)
2160 Hz
- (d)
72 Hz
The velocity of sound waves in an ideal gas at temperatures T1 (K) and T2 (K) are respectively v1 and v2. The rms velocities of gas molecules at these two temperatures are w1 and w2 respectively; then:
- (a)
\(\frac{v_{1}}{v_{2}}=\frac{w_1}{w_2}\)
- (b)
\(\frac{v_1}{v_2}=\sqrt{\gamma}(\frac{w_1}{w_2})\)
- (c)
\(\frac{v_1}{v_2}=\sqrt{\frac{\gamma}{3}}(\frac{w_1}{w_2})\)
- (d)
\(\frac{v_1}{v_2}=\sqrt{\frac{w_1}{w_2}}\)
\(v_1=\sqrt{\frac{\gamma RT_1}{M}}, v_2=\sqrt{\frac{\gamma RT_2}{M}}\)
\(w_1=\sqrt{\frac{3RT_1}{M}}; w_2=\sqrt{\frac{3RT_2}{M}}\)
\(\frac{v_1}{v_2}=\frac{w_1}{w_2}=\sqrt{\frac{T_1}{T_2}}\)
A heavy rope is suspended from a rigid support. A wave pulse is set up at the lower end; then:
- (a)
the pulse will travel with uniform speed
- (b)
the pulse will travel with increasing speed
- (c)
the pulse will travel with decreasing speed
- (d)
the pulse cannot travel through the rope
Due to weight of the rope, the tension will be increasing along the rope from the lower end to the upper end. Hence, the pulse will travel with increasing speed as v=\(\sqrt{\frac{T}{m}}\)
It takes 2.0 seconds for a sound wave to travel between two fixed points when the day temperature is 10° C. If the temperature rises to 300 C the sound wave travels between the same fixed points in:
- (a)
1.9 sec
- (b)
2.0 sec
- (c)
2.1 sec
- (d)
2.2 sec
\(t=\frac{s}{v}\)
\(\frac{t_1}{t_2}=\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\)
\(\frac{2}{t_2}=\sqrt{\frac{303}{283}}\) or t2=1.9sec
The intensity of a plane progressive wave of frequency 1000 Hz is 10-10 watt per metre 2 . Given that the speed of sound is 330 m/s and density of air is 1.293kg/m3, then the maximum change in pressure (in N/m2) is:
- (a)
3 x 10-4
- (b)
3 x 10-5
- (c)
3 x 10-3
- (d)
3 x 10-2
In a sports meet, the timing of a 200 metre straight dash is recorded at the finish point by starting an accurate stop watch on hearing the sound of starting gun fired at the starting point. The time recorded will be more accurate:
- (a)
in winter
- (b)
in summer
- (c)
in all seasons
- (d)
none of these
Time recorded in summer is more accurate. The velocity of sound is directly proportional to the square root of absolute temperature. Hence, the sound of the gun fired at the starting point will reach the finishing point quicker in summer than in winter. The lapse of time due to the time taken by the sound in reaching the finish point will be less in summer and hence the time recorded will be more accurate in summer than in winter.
If sound waves can be assumed to be diffracted which of the following objects will diffract sound waves in air from a 384 Hz tuning fork?
- (a)
A sphere of radius 10 m
- (b)
A sphere of radius 1m
- (c)
A sphere of radius 1 mm
- (d)
A sphere of radius 1 cm
The phase difference between two waves, represented by:
y1=10-6 sin[100t+(x/50)+0.5]m
y2=10-6 cos[100t+(x/50)]m
where x is expressed in metres and t is expressed in seconds, is approximately:
- (a)
1.07 rad
- (b)
2.07 rad
- (c)
0.5 rad
- (d)
1.5 rad
A line source emits a cylindrical wave. If the medium absorbs no energy, the amplitude will vary with distance r from the source as proportional to:
- (a)
r-1
- (b)
r-2
- (c)
r-1/2
- (d)
r1/2
The energy is inversely proportional to the square of distance. Hence, the amplitude is inversely proportional to the distance.
When a wave travels in a medium, the particle displacement is given by: y=a sin2π(bt-cx) where a, b and c are constants. The maximum particle velocity will be twice the wave velocity if:
- (a)
\(c=\frac{1}{\pi a}\)
- (b)
c=πa
- (c)
b=ac
- (d)
b=\(\frac{1}{ac}\)
- (e)
a=bc
Equation of the harmonic progressive wave given by:
y=a sin 2π(bt-cx)
Here, 2πv=ω=2πb
or v=b
k=\(\frac{2\pi}{\lambda}=2\pi c\)
or \(\frac{1}{\lambda}\)=c
(Here c is the symbol given for \(\frac{1}{\lambda}\) and not the velocity)
ஃ Velocity of the wave=vλ=\(b\frac{1}{c}=\frac{b}{c}\)
\(\frac{dy}{dt}\)=a2πb cos2π(bt-cx) =aω cos(ωt-kx)
Maximum particle velocity=aω=a 2πb=2π ab. Given this is equal to \(2\times\frac{b}{c}\)
i.e., \(2\pi a=\frac{2}{c}\) or c=\(\frac{1}{\pi a}\)
The frequency of a tuning fork with an amplitude A = 1cm is 250 Hz. The maximum velocity of any particle in air is equal to:
- (a)
\(\frac{5}{\pi}m/sec\)
- (b)
5π m/sec
- (c)
\(\frac{3.30}{\pi}m/sec\)
- (d)
none of these
What is the maximum possible sound level of sound waves in air? Given that density of air = 1.3 kg/m 3, u = 332 m/s and atmospheric pressure = 1.01 x 105 N/m 2:
- (a)
120 dB
- (b)
60 dB
- (c)
190 dB
- (d)
50 dB
\(I=\frac { { P }_{ 0 }^{ 2 } }{ 2\rho v } =\frac { { ({ 1.01\times 10 }^{ 5 }) }^{ 2 } }{ 2\times 1.3\times 332 } \)
\(=1.18\times { 10 }^{ 7 }{ W/m }^{ 2 }\)
\(\therefore SL=10\log { \frac { I }{ { I }_{ 0 } } } =10\log { \frac { { 10 }^{ 7 } }{ { 10 }^{ -12 } } =190dB } \)
The power of sound from the speaker of a radio is 20 mW. By turning the knob of volume control the power of sound is increased to 400 mW. What is the power increase as compared to the original power?
- (a)
1.3 dB
- (b)
3.1 dB
- (c)
13 dB
- (d)
30.1 dB
As P∝I,
\(\therefore { SL }_{ 2 }-{ SL }_{ 1 }=10\log { \left( \frac { { I }_{ 2 } }{ { I }_{ 1 } } \right) } =10\log { \left( \frac { { P }_{ 2 } }{ { P }_{ 1 } } \right) } \)
\(=10\log { \left( \frac { 400 }{ 20 } \right) } =10\log { 20 } \simeq 13dB\)
The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 0.5π. The wave velocity is:
- (a)
144 m/s
- (b)
256 m/s
- (c)
384 m/s
- (d)
720 m/s
\(\Delta x=0.8m,\quad \quad n=120Hz,\quad \Delta \phi =0.5\pi \)
\(\Delta \phi =\frac { 2\pi }{ \lambda } .\Delta x\)
\(or\quad \lambda =2\pi .\frac { \Delta x }{ \Delta \phi } \)
\(=\frac { 2\pi }{ 0.5\pi } .0.8=3.2m\)
Wave velocity, v = nλ = 120 x 3.2 = 384 m/s.
A sings with a frequency (n) and B sings with a frequency (118) that of A. If the energy remains the same and the amplitude of A is a then amplitude of B is:
- (a)
a
- (b)
2a
- (c)
8a
- (d)
16a
\({ n }_{ A }=n,{ n }_{ B }=\frac { n }{ 8 } ,{ a }_{ A }=a\)
\(E=\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } m{ (a\omega ) }^{ 2 }=\frac { 1 }{ 2 } { ma }^{ 2 }{ \omega }^{ 2 }\)
\(i.e.,\quad E\alpha { a }^{ 2 }\times { (2\pi n) }^{ 2 }orE\alpha { a }^{ 2 }{ n }^{ 2 }\)
\(\therefore \frac { { E }_{ A } }{ { E }_{ B } } ={ \left( \frac { { a }_{ A } }{ { a }_{ B } } \right) }^{ 2 }\times { \left( \frac { { n }_{ A } }{ { n }_{ B } } \right) }^{ 2 }\)
As EA =EB,hence
\({ \left( \frac { { a }_{ A } }{ { a }_{ B } } \right) }^{ 2 }={ \left( \frac { { n }_{ B } }{ { n }_{ A } } \right) }^{ 2 }\)
or \(\frac { { a }_{ A } }{ { a }_{ B } } =\frac { { n }_{ B } }{ { n }_{ A } } =\frac { n/8 }{ n } =\frac { 1 }{ 8 } \)
or\({ a }_{ B }=8{ a }_{ B }=8a\)
The SHM of a particle is given by: X (t) = 5 cos\(\left( 2\pi t+\frac { \pi }{ 4 } \right) \). (in MKS units). Calculate the displacement and magnitude of acceleration of the particle at t = 1.5seconds
- (a)
-3.0m,100m/s2
- (b)
+2.54m,200m/s2
- (c)
-3.54m,140m/s2
- (d)
+3.55m,120m/s2
The given equation of SHM is,
X(t)=5cos\(\left( 2\pi t+\frac { \pi }{ 4 } \right) \)
Compare the given equation with standard equation of SHM
X (I) = A cos (ωt +φ )
We get, ω = 2πsec-1
At t =1.5 sec:
(i) Displacement:
X (t) = 5 cos\(\left( 2\pi \times 1.5+\frac { \pi }{ 4 } \right) =5cos\left( 3\pi +\frac { \pi }{ 4 } \right) \)
\(=-5cos\frac { \pi }{ 4 } =-5\times 0.707m=-3.54m\)
(ii) Acceleration:
a =-ω2 x Displacement
= - (2 π s-1)2 X (- 3.54 m)
= 140 rns-2.
The ratio of amplitudes of two simple harmonic motions represented by the equations Y1 = 5sin\(\left( 2\pi t+\frac { \pi }{ 4 } \right) \)Y2 = 2.√2(sin 2πt+cosπt) is
- (a)
1 : 1
- (b)
1 : 2
- (c)
5 : 2
- (d)
5 : 4
- (e)
2 : 5
Given that y1 = 5 sin\(\left( 2\pi t+\frac { \pi }{ 4 } \right) \)
Comparing with standard equation, we get A1= 5
Further, \({ y }_{ 2 }=2\sqrt { 2 } (sin2\pi t+cos2\pi t)\)
or, \({ y }_{ 2 }=2\sqrt { 2 } \times \sqrt { 2 } \left[ \frac { 1 }{ \sqrt { 2 } } sin2\pi +\frac { 1 }{ \sqrt { 2 } } cos2\pi t \right] \)
\(=4\left[ sin2\pi t.cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } .cos2\pi t \right] \)
=4sin\(\left( 2\pi t+\frac { \pi }{ 4 } \right) \)
Hence,A2=4
\(\therefore \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 5 }{ 4 } \)
The bulk modulus of a liquid of density 8000 kg m-3 is 2 x 109Nm-2. The speed of sound in that liquid is (in m s -1) :
- (a)
200
- (b)
250
- (c)
100
- (d)
350
- (e)
500
The speed of sound in liquid
\(v=\sqrt { \frac { B }{ \rho } } \)
Here, it is given that, B = 2 x 109N / m2 and p = 8000 kg/m3
\(\therefore v=\sqrt { \frac { 2\times { 10 }^{ 9 } }{ 8000 } } =\sqrt { \frac { 1 }{ 4 } \times { 10 }^{ 6 } } \)
\(=\frac { 1 }{ 2 } \times { 10 }^{ 3 }\)
= 500 ms-1
Hence, correct answer is (e).
A travelling acoustic wave of frequency 500 Hz is moving along the positive x-direction with a velocity of 300 m s -1. The phase difference between two points X1 and X2 is 60°. Then the minimum separation between the two points is :
- (a)
1 mm
- (b)
1 cm
- (c)
10 cm
- (d)
1 m
The relation between phase difference and path difference is given by
\(\Delta \phi =\frac { 2\pi }{ \lambda } \Delta x\)
Here, \(\Delta \phi ={ 60 }^{ o }=\frac { \pi }{ 3 } ,v=300{ ms }^{ -1 }and\quad v=500Hz\)
\(\therefore \frac { \pi }{ 3 } =\frac { 2\pi }{ \left( \frac { 300 }{ 500 } \right) } \Delta x\quad \quad [\because v=v\lambda ]\)
\(\therefore \Delta x=\frac { 3 }{ 2\times 5\times 3 } =\frac { 1 }{ 10 } m=10cm\)
Hence, correct answer is (c).
If wave y=A cos(ωt+kx) is moving along x-axis, the shape of pulse at t = 0 and t = 2 s:
- (a)
are different
- (b)
are same
- (c)
may not be same
- (d)
none of these
At t = 0 and t = 2 s, the shapes of y-x graphs are same.