Practice free Waves (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Two sound waves are represented by y = a Sin(ùt-kx) and y = a Cos(ùt-kx). The phase difference between the waves in water is ……..
As $sin(90 \pm \theta ) = cos \theta $ The phase difference between the two waves is $ \pi /2 $
A string of linear density 0.2 kg/m is stretched with a force of 500 N. A transverse wave of length 4.0 mand amplitude 1/l meter is travelling along the string. The speed of the wave is………….m/s.
$ \nu = \sqrt T = \sqrt { 500 \over 0.2 } = 50 ms^{-1} $
Two wires made up of same material are of equal lengths but their radii are in the ratio 1:2. On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is
$ Here , \rho_1 = \rho_2 , { r_1 \over r_2 } = {1 \over 2 } , T_1 = T_2 $ $ f_1 = { 1 \over 2lr_1 } \sqrt { T_1 \over \pi \rho_1 } , f_2 = { 1 \over 2lr_2 } \sqrt { T_2 \over \pi \rho_2 } , $ $ \therefore { f_1 \over f_2 } = {r_1 \over r_2 } = { 2 \over 1 } $
The tension in a wire is decreased by 19%, then the percentage decrease in frequency will be ………
$ { f_2 \over f_1 } = sqrt { T_2 \over T_1 } = \sqrt { 81 \over 100 } = { 9 \over 10 } $ $ \therefore { f_1 - f_2 \over f_1 } \times 100 = 10 \%$
An open organ pipe has fundamental frequency 100 hz. What frequency will be produced if its one end is closed?
When one end is closed $f_1 = {100 \over 2 } = 50 Hz $ $ f_2 = 3f_1 =150 Hz , f_3 = 5f_1 =250Hz and so on...$
A closed organ pipe has fundamental frequency 100 hz. What frequencies will be produced if its other end is also opened?
When other end of pipe is opened, its fundamental frequency becomes 200Hz. The overtone have frequencies 400, 600, 800 Hz..
A column of air of length 50 cm resonates with a stretched string of length 40 cm. The length of the same air column which will resonate with 60 cm of the same string at the same tension is ……..
$ As , { l_2 \over 2l }= { l_2' \over l_1'} \Rightarrow { 60 \over 40 } = { l_2' \over 50 } = l_2' = 75cm $
Two forks A and B when sounded together produce 4 beats/s. The fork A is in unison with 30 cm length of a sonometer wire and B is in unison with 25 cm length of the same wire at the same tension. The frequencies of the fork are
$ { f_2 \over f_1 } = { l_2 \over l_1 } = { 25 \over 30 } = { 5 \over 6 } $ $ f_2 - f_1 = 4 on solving we get f_2 = 24 Hz $ $ \therefore f_1 = 20 Hz $
A tuning fork of frequency 200 hz is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by 1 % is
$ { f_2 \over f_1 } = \sqrt { 101 \over 100 } = \left( 1 + { 1 \over 100 } \right) ^ {1 /2 } = 1 + {1 /200} $ $ \therefore f_2 = f_1 + { f_1 \over 200} $ $ \therefore numbers of be ab s^{-1} = f_2 -f_1 = {f_1 \over 200} = 1 $
A bus is moving with a velocity of 5 m/s towards a huge wall. The driver sounds a horn of frequency 165 hz. If the speed of sound in air is 335 m/s, the number of beats heard per second by the passengers in the bus will be …….
$ { f_L \over f_S} = { \nu + \nu_L \over \nu + \nu_S } $ $ here \nu_L = + 5 ms^{-1} , \nu_s = -5 ms^{-1} , f_s = 165 Hz $ $ \therefore f_L =170 Hz \therefore Number of be ab s^{-1} = 170 -165 = 5 $
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