IISER Physics - Waves
Exam Duration: 45 Mins Total Questions : 30
Choose the INCORRECT statement
- (a)
Audibility range of a human ear is 20 Hz to 20 kHz
- (b)
Linear distance between the successive points which are in the same phase is half of the length of a wave
- (c)
The temperature at which velocity of sound in air is double its velocity at \({ 0 }^{ \circ }C\) is \(819^{ \circ }C\)
- (d)
The velocity of sound increases with increase of humidity
Linear distance between the successive points which are in the same phase is wavelength of the wave.
Sound travels fastest in
- (a)
hydrogen
- (b)
vacuum
- (c)
water
- (d)
steel
Velocity of sound in a medium is inversely proportional to square-root of density of the medium. The density of steel is greatest out of the hydrogen, water and vacuum. Sound does not travel in vacuum.
Sound waves cannot be
- (a)
refracted
- (b)
polarised
- (c)
scattered
- (d)
used to produce hologram
Holography is exhibited by all types of waves-longitudinal as well as transverse.
Choose the CORRECT statement
- (a)
Longitudinal waves are possible in solids, liquids and gases
- (b)
Longitudinal waves are possible only in solids and liquids
- (c)
Both transverse and longitudinal waves are possible in solids, liquids and gases
- (d)
Transverse waves are not possible in vacuum
A pulse along a stretched string travels at a speed of 5 ms-1 and frequency of 20 Hz. What is the phase difference in radians between two points situated 2.5 am apart?
- (a)
\(\pi /4\)
- (b)
\(\pi /5\)
- (c)
\(\pi /6\)
- (d)
\(\pi /8\)
Wavelength, \(\lambda =\frac { \vartheta }{ \upsilon } =\frac { 5 }{ 20 } =0.25m\)
If \(\phi \) is phase difference and \(\delta \) , the path difference,
\(\frac { 2\pi }{ \phi } =\frac { \lambda }{ \delta } \)
\(\phi =\frac { 2\pi }{ \lambda } \delta \)
Given \(\delta \) =2.5cm=2.5x10-2m
\(\phi =\frac { 2\pi }{ 0.25 } \times 2.5\times { 10 }^{ -2 }=\frac { \pi }{ 5 } \)rad.
Two sound waves of the same frequency have respective amplitudes of 3 units and 1 unit and are travelling in opposite directions in the same straight line. At a particular place in that line, the resultant wave will vary in loudness. The ratio of maximum loudness to minimum loudness is
- (a)
9 : 1
- (b)
6 : 1
- (c)
9 : 2
- (d)
4 : 1
Maximum amplitude=3+1=4 units
Minimum amplitude=3-1=2 units
Required ratio=\(\frac { \left( 4 \right) ^{ 2 } }{ \left( 2 \right) ^{ 2 } } =\frac { 4 }{ 1 } .\)
One of the similarities between sound and light is that both
- (a)
are transverse waves
- (b)
can show holographic effects
- (c)
can pass through vacuum
- (d)
travel with constant speed in free space
Both sound and light exhibit holographic effect which is based on interference effects.
In the production of beats by two progressive waves of nearly the same frequency,
- (a)
the frequency of the beats is a function of time
- (b)
the frequency of the beats depends on the relative position of the listener
- (c)
the frequency of beats depends on the relative velocity between the source and listener
- (d)
the frequency of the beats can heard more distinctly if the frequency difference between the component waves is large
Alternative (c) can be explained on the basis of Doppler principle in sound.
Source of sound and observer are moving with the same velocity along the positive direction of x-axis. The medium is moving along the negative direction of x-axis. Let \({ \vartheta }_{ s },{ \quad \vartheta }_{ 0 }\) and w be velocities of source, observer and medium respectively, and \(\nu \quad and\quad { \nu }^{ \prime }\) be the frequency of sound source and the apparent frequency heard by the observer respectively. Then,
- (a)
\({ \nu }^{ \prime }\quad =\quad \nu \)
- (b)
\({ \nu }^{ \prime }\quad =\quad \left( \frac { { \nu }_{ s }-{ \nu }_{ 0 }-w }{ { \nu }_{ s } } \right) \nu \)
- (c)
\({ \nu }^{ \prime }\quad =\quad \left( \frac { { \nu }_{ s }-w }{ { \nu }_{ s } } \right) \nu \)
- (d)
\({ \nu }^{ \prime }\quad =\quad \left( \frac { { \nu }_{ s }+w }{ { \nu }_{ s } } \right) \nu \)
Relative velocity between sound source and observer is zero.
Velocity of sound relative to medium = \({ \vartheta }_{ s }-w\) A change of velocity of the wave changes its frequency. N.B. If the velocity of the medium is in the direction of the sound, apparent frequency is greater and vice versa. In such
a case, \(\upsilon '=\frac { { \vartheta }_{ s }+w }{ { \vartheta }_{ s } } \upsilon .\)
The water waves are approaching obliquely a beach. They are found to bend usually towards the short line as they move closer. This depicts the phenomenon of
- (a)
reflection
- (b)
refraction
- (c)
interference
- (d)
diffraction
Water waves approaching a beach at an angle are often noticed bending towards the shortline as they move closer. This is an example of diffraction of waves.
The velocity of sound in carbon dioxide is less than in hydrogen because
- (a)
carbon dioxide has more degrees of freedom
- (b)
carbon dioxide is soluble in water
- (c)
carbon dioxide is heavier than hydrogen
- (d)
carbon dioxide is a compound and hydrogen is an element
Velocity of sound varies inversely with the square root of the density of gas.
Velocity of sound is inversely proportional to the square root of molecular weight of a gas.
The equation
\(y\quad =\quad a\quad sin\quad \frac { 2\pi }{ \lambda } (\vartheta t+x+{ \phi }_{ 0 })\)
represents
- (a)
a wave, longitudinal or transverse, travelling along minus x-axis
- (b)
a wave, longitudinal or transverse, travelling along positive x-axis
- (c)
a wave, longitudinal or transverse, travelling along y-axis
- (d)
a transverse wave travelling along minus x-axis
Two travelling waves of sound are found to have sinusoidal waveform but different wavelength and different amplitude. They will have
- (a)
same pitch but different intensity
- (b)
same pitch but different quality
- (c)
same quality but different intensity
- (d)
different quality and intensity
Two sound waves having different amplitudes have different intensity because intensity of a wave is proportional to (amplitude).
Two sound waves having sinusoidal waveform have the same quality (also known as timbre of sound).
A closed and an open organ pipe are sounded to the same frequency. The ratio of their lengths is
- (a)
1 : 1
- (b)
1 : 2
- (c)
2 : 1
- (d)
1 : 4
A closed organ pipe has length \({ l }_{ 1 }=\lambda /4\)
An open organ pipe has length \({ l }_{ 2 }=\lambda /2\)
\({ l }_{ 1 }:{ l }_{ 2 }=\frac { \lambda }{ 4 } :\frac { \lambda }{ 2 } =1:2\)
The quality of a tone
- (a)
depends on the overtones
- (b)
depends on the notes
- (c)
decrease with pitch
- (d)
increase with loudness
piezo-electric effect is exibited by
- (a)
diamond
- (b)
glass
- (c)
quartz
- (d)
nickel
The magnitude of threshold of hearing, persistence of hearing and persistence of vision are respectively
- (a)
\({ 10 }^{ -6 }\quad W{ m }^{ -2 },\quad \frac { 1 }{ 5 } s\quad and\quad \frac { 1 }{ 10 } s\)
- (b)
\({ 10 }^{ -10 }\quad W{ m }^{ -2 },\quad \frac { 1 }{ 10 } s\quad and\quad \frac { 1 }{ 25 } s\)
- (c)
\({ 10 }^{ -16 }\quad W{ m }^{ -2 },\quad \frac { 1 }{ 16 } s\quad and\quad \frac { 1 }{ 10 } s\)
- (d)
\({ 10 }^{ -12 }\quad W{ m }^{ -2 },\quad \frac { 1 }{ 10 } s\quad and\quad \frac { 1 }{ 16 } s\)
Any note and its major tone bear ratio
- (a)
16 : 5
- (b)
15 : 16
- (c)
8 : 9
- (d)
9 : 8
A sonometer of fixed length is to produce one octave lower than before by varying the tension. The fractional change in tension required for this purpose will be
- (a)
1/4
- (b)
3/4
- (c)
1/2
- (d)
3/4
Initial tension=T1 and initial frequency=\({ \upsilon }_{ 1 }\)
Final tension=T2 and initial frequency=\({ \upsilon }_{ 2 }\)
Given that \(\frac { { \upsilon }_{ 2 } }{ { \upsilon }_{ 1 } } =\frac { 1 }{ 2 } \)
and \(\frac { { \upsilon }_{ 2 } }{ { \upsilon }_{ 1 } } =\sqrt { \frac { { T }_{ 2 } }{ { T }_{ 1 } } } \)
\(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1 }{ 4 } \)
\(I=\frac { { T }_{ 2 } }{ { T }_{ 1 } } =1-\frac { 1 }{ 4 } \) which is \(\frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } =\frac { 3 }{ 4 } \)
Two sources of sound of frequencies \({ \nu }_{ 1 }\quad and\quad { \nu }_{ 2 }\) superpose each other to give rise to the formation of beats. In beats,
- (a)
the phase at a point remains constant
- (b)
the amplitude at a point remains constant
- (c)
the amplitude at a point changes at the rate of \({ (\nu }_{ 1 }\quad -\quad { \nu }_{ 2 })\)
- (d)
the phase at a point changes at the rate of \(\frac { 1 }{ 2 } { (\nu }_{ 1 }\quad -\quad { \nu }_{ 2 })\)
A pipe closed at one end resonates to its fundamental frequency of 800 Hz. Which frequencies from A, B, C and D can the pipe also resonate to?
A. 1600 Hz B. 2400 Hz C. 3200 Hz D. 4000 Hz
- (a)
if A and B are correct
- (b)
if B and C are correct
- (c)
if B and C are correct
- (d)
if A, B, C and D are correct
For a closed pipe, only odd harmonics are possible. The frequencies ratio are
\({ \upsilon }_{ 1 }:{ \upsilon }_{ 2 }:{ \upsilon }_{ 3 }\) etc.=1:3:5 etc.
Thus, possible frequencies 800x3, 800x5 which are 2,400 Hz and 4000 Hz.
The mechanical waves have in common with electromagnetic waves the essential feature of
- (a)
energy propagation
- (b)
momentum propagation
- (c)
propagation through any meterial substance-solid, lliquid and gas
- (d)
both energy and momentum propagation
Both mechanical and electromagnetic waves can propagate through any Material substance-solid, liquid and gas. In the case of electromagnetic waves, there is no need even for the existence of a medium to sustain the wave motion. However, they have in common with mechanical waves the essential feature of energy and momentum.
A simple harmonic wave propagates along a taut string. The wavelength of the wave is 0.75 m and its time period is 5.30 ms. The average power transmitted along the string is 1.24 kW, then the total energy per wavelength in the wave is
- (a)
6.57 J
- (b)
0.23 X 106 J
- (c)
1.55 X 103 J
- (d)
NONE OF THE ABOVE
If Pav is the average power transmitted along the string and T is time period of the wave, then the total energy per wavelength in the wave is
\(E={ \rho }_{ av }T\)
\(=\left( 1.24\times { 10 }^{ 3 }W \right) \left( 5.30\times { 10 }^{ -3 }S \right)\)
=6.57J
By using special laboratory apparataus, it is possible to produce sound waves with wavelength in air as small as, comparable with wavelength of
- (a)
infrared light
- (b)
visible light
- (c)
ultraviolet light
- (d)
x-rays
In the laboratory it is possible to produce sound waves with frequencies of upto to about 1 GHz =109 Hz.Such waves have wavelength in air, \(\lambda =\frac { 340\quad { ms }^{ -1 } }{ { 10 }^{ 9 }Hz } \cong 340nm\)
A long thin copper rod is given a sharp compressional blow at one end. The sound of the blow, travelling through air at \({ 0 }^{ \circ }C\), reaches the opposite end of the rod 6.84 ms later than the sound transmitted through the rod. The velocity of sound in copper rod is \(\vartheta { C }_{ u }\quad =\quad 3.750\quad km\quad { s }^{ -1 }\) and velocity of sound in the air at \({ 0 }^{ \circ }C\) is \(\vartheta \quad =\quad 330\quad m{ s }^{ -1 }\). The length of the rod is
- (a)
2.475 km
- (b)
2.475 m
- (c)
2.475 cm
- (d)
NONE OF THE ABOVE
Let I be the length of the rod
\(\frac { l }{ \vartheta { C }_{ u } } =t{ C }_{ u }\quad and\quad \frac { l }{ \vartheta } =t\)
\(\frac { l }{ \vartheta } -\frac { l }{ \vartheta { C }_{ u } } =t-t{ C }_{ u }\)
\(l=\left( \frac { \vartheta { C }_{ u }.\vartheta }{ \vartheta { C }_{ u }-\vartheta } \right) \left( t-t{ C }_{ u } \right) \)
\(=\frac { 3750\times 330 }{ 3750-330 } \times 684\times { 10 }^{ -3 }\)
=2.475m
A tuning fork whose natural frequency is 440 Hz is placed immediately above a tube that contains water. The water is slowly drained from the tube while the vibrations of the tuning fork excite the air in the tube above the water level. It is found that the sound is enhanced when the air column has a length 56.6 cm and next when it has a length of 96.6 cm. The temperature of air is
\([velocity\quad of\quad sound\quad at\quad { 0 }^{ \circ }C\quad is\quad { \vartheta }_{ 0 }\quad =\quad 320\quad m{ s }^{ -1 }]\)
- (a)
\({ 330.3 }^{ \circ }C\)
- (b)
\({ 300.3 }^{ \circ }C\)
- (c)
\({ 57.3 }^{ \circ }C\)
- (d)
\({ 27.3 }^{ \circ }C\)
\(\frac { \lambda }{ 2 } \)=96.6-56.6=40.0cm
\(\lambda \)=2x40=80cm=0.80m
Velocity of sound at temperature to C is
\(\vartheta _{ t }=\upsilon \lambda =440\times 0.80=352{ ms }^{ -1 }\)
\(\frac { \vartheta _{ t } }{ \vartheta _{ 0 } } =\sqrt { \frac { T }{ { T }_{ 0 } } } or\frac { 352 }{ 320 } =\sqrt { \frac { T }{ 273 } } \)
\(1.1=\sqrt { \frac { T }{ 273 } } or\quad T=273\times 1.21=330.3K\)
\({ t }^{ o }C=330.3-273.0=57.3^{ o }{ C }\)
In stationary wave the strain is
- (a)
maximum at nodes
- (b)
maximum at antinodes
- (c)
constant throughout
- (d)
NONE OF THE ABOVE
A uniform rope of mass 0.1 kg and length 2.45 m hangs from the ceiling. Find the speed transverse waves at a point 0.5 m distant from the lower end.
- (a)
1.22 ms-1
- (b)
2.21 ms-1
- (c)
22.1 ms-1
- (d)
12.2 ms-1
Speed of transverse waves is \(\vartheta =\sqrt { \frac { T }{ m } } \)
Tension T = ml g of where m is mass per unit length
\(\vartheta =\sqrt { \frac { mlg }{ m } } =\sqrt { lg } =\sqrt { 0.5\times 9.8 } =2.21{ ms }^{ -1 }\)
A simple rod clamped in the middle can be set into longitudinal vibration. What kind of vibrations does it produce when bent to form a turning fork?
- (a)
Transverse
- (b)
Longitudinal
- (c)
Both transverse and longitudinal
- (d)
Transverse or longitudinal depending upon the medium
A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A record a frequancy 5.5. kHz while the train apporoaches the siren. During his return journy in a different train B he record a frequency 6.0 kHz while approaching the same siren. The radio of the velocity of train B to that of train A is.
- (a)
\(242\over 256\)
- (b)
2
- (c)
\(5\over 6\)
- (d)
\(11\over 6\)
Let \({ \vartheta }_{ L }^{ / }\) be the velocity of train A
\({ \upsilon }'=\frac { \vartheta +{ \vartheta }_{ L }^{ / } }{ \vartheta } \)
\({ \upsilon }'\)=5.5kHz, \(\upsilon \)=5kHz
\(\frac { { \upsilon }' }{ \upsilon } =1+\frac { { \vartheta }_{ L }^{ / } }{ \vartheta } or\frac { { \vartheta }_{ L }^{ / } }{ \vartheta } =\frac { { \upsilon }' }{ \upsilon } -1=\frac { 5.5 }{ 5 } -1=0.1\)
Similarly, \(\frac { { \vartheta }_{ L }^{ " } }{ \vartheta } =\frac { { \upsilon }'' }{ \upsilon } -1=\frac { 6 }{ 5 } -1=0.2\)
\(\frac { { \vartheta }_{ L }^{ " } }{ { \vartheta }_{ L }^{ ' } } =\frac { 0.2 }{ 0.1 } =2\)