Practice free Waves (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The wave number for a wave having wavelength 0.005 m is…….$m^{– 1}$ .
$ wave number = { 1 \over \lambda } = { 1 \over 0.005 } = 200 m^{-1} $
An listener is moving towards a stationary source of sound with a speed 1/4 times the speed of sound. What will be the percentage increase in the frequency of sound heard by the listener?
Frequency heard by the listener $ f_L = \left ( { \nu + \nu_L \over \nu } \right) f_s $ $ ( \therefore \nu_s = 0 ) $ $ \therefore { f_L \over f_s } = { \nu + \nu_2 \over \nu } = { \nu + { \nu \over 4 } \over \nu } = { 5 \over 4 } $ $ \therefore \% increase = { f_L - f_s \over f_s } \times 100 = \left( { 5 - 4 \over 4 } \right) \times 100 = 25 \% $
When the resonance tube experiment, to measure speed of sound is performed in winter, the first harmonic is obtained for 16 cm length of air column. If the same experiment is performed in summer, the second harmonic is obtained for x length of air column. Then
$ From \nu = \sqrt { \gamma RT \over M } , \nu \alpha \sqrt T $ In summer, velocity increases & hence decreases and so L increases. The length of 2nd halmonics $ x = 3L_1 = 3 \times 16 = 48 cm $ In summer, velocity being more, $ x \gt 3L_1$ $ \therefore x \gt 48 $
What should be the speed of a source of sound moving towards a stationary listener, so that the frequency of sound heard by the listener is double the frequency of sound produced by the source? { Speed of sound wave is v }
$ In f_L = \left( { \nu + \nu_L \over \nu + \nu_s } \right) f_s $ $ putting \nu_L = 0 , f_L = 2 f_s , \nu = \nu, \nu_s = - \nu_s $ $ 2 f_s = \left( { \nu \over \nu - \nu_s } \right) f_s \Rightarrow 2 \nu_s = \nu \therefore \nu_s = { \nu \over 2 } $
If the listener and the source of sound moves along the same direction with the same speed, then……..
$ { f_L \over f_s } = { \nu + \nu_L \over \nu + \nu_s } or { f_L \over f_s } = { \nu - nu_L \over \nu - \nu_s } $ $ but , \nu_L = \nu_s $ $ \therefore { f_L \over f_s } = 1 $
A wire of length 10 mand mass 3 kg is suspended from a rigid support. The wire has uniform cross sectional area. Now a block of mass 1 kg is suspended at the free end of the wire and a wave having wavelength 0.05 m is produced at the lower end of the wire. What will be the wavelength of this wave when it reached the upper end of the wire?
Since the rope is heavy, the tension at the lower end & top end of the rope will be different. Mass of rope $ m_2 $ = 3kg Mass of block $ m_1 $ = 1 kg $ \therefore tension at the lower end T_1 = m_1 g = 1 g N and at the upper end in T_2 = (m_1 + m_2 ) g = 4 g N $ Now speed of wave in rope $ \nu = \sqrt T \Rightarrow f \lambda = \sqrt T $ $ \therefore \lambda = \sqrt T ( \therefore f , \mu are constants ) $ $ \therefore Wave length at lower end and \lambda_1 = \sqrt T_1 and at the upper end \lambda_2 = \sqrt T_ 2$ $ \therefore { \lambda_2 \over \lambda_1 } = \sqrt { T_2 \over T_1 } \Rightarrow \lambda_2 = \sqrt { T_2 \over T_1 } = \lambda_1 = \lambda_1 = 0.1 m $
If the mass of 1 mole of air is $29 x 10^{– 3} kg$, then the speed of sound in it at STP is……..( ã=7/5). ${ T = 273 K, P = 1.01 x 10^5 Pa }$
$ Speed of sound = \sqrt { \gamma P \over \rho } $ $ \rho = { mass of 1 mole air \over volume of 1 mole air } = { 29 \times 10^{-3} kg \over 22.4 \times 10^{-3} m^3 } = 1.3 $ $ \therefore speed = \sqrt { { 7 \over 5} \times { 1.01 \times 10^5 \over 1.3 } } = 330 ms^{-1} $
A wave travelling along a string is described by y = 0.005Sin(40x – 2t) in SI units. The wavelength and frequency of the wave are………
$ From the phase angle ( 40 -2t ) , we get k =40 \, OR\, { 2 \pi \over \lambda } = 40 \Rightarrow \lambda = { \pi \over 20 } $ $ and \omega = 2 \, OR \, 2 \pi f = 2 \Rightarrow f = \pi^{-1} Hz $
Two sitar strings A and B playing the note “Dha†are slightly out of time and produce beats of frequency 5 hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 hz. What is the original frequency of B if the frequency of A is 427 hz?
Increase in tension of string increases its frequency. If the original frequency of $B(f_B)$ were greater than that of $A(f_A)$, further the increase in $f_B$ should have resulted in increase in the beat frequency. But the beat frequency is found to decrease. This shows that $f_A-f_B = 5 Hz$ and $f_A=427 Hz$, we get $f_B = 422 Hz$
A rocket is moving at a speed of 130 m/s towards a stationary target. While moving, it emits a wave of frequency 800 hz. Calculate the frequency of the sound as detected by the target. ( Speed of wave = 330 m/s)
$ f_L = \left( { \nu - \nu_L \over \nu - \nu_s } \right) f_s = \left[ { 330 - 0 \over 330 -130 } \right] \times 800 = 1320 Hz $
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