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A wave y = aSin(ùt – kx ) on a string meets with another wave producing a node at x= 0. Then the equation of the unknown wave is ………

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Explanation

Stationary wave : Y = a sin (wt-kx) + a sin(wt+kx) When x = 0, $Y \neq 0$. The option is not acceptable consider option (b) stationary wave : Y = a sin (wt-kx) - a sin(wt+kx) At x = 0, Y = 0. This option holds good. Option (c) gives Y = 2a sin(wt - kx) At x = 0, $Y \neq 0$ Option (d) gives Y = 0. Hence option (b) holds good

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A tuning fork of known frequency 256 hz makes 5 beats per second with the vibrating string of a piano. The beats frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increase in the tension was hz.

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Explanation

The possible frequency of piano are (256 + 5)Hz and (256 - 5)Hz. For a piano string $ \nu = { 1 \over 2l } \sqrt {T} $ When tension T increases v increases. (i) If 261 Hz increases, beats / second increase. This is not given. (ii) If 251 Hz increases due to tension, beats / second decrease. This is given.

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An observer moves towards a stationary source of sound with a velocity one – fifth the velocity of sound. What is the percentage increase in the apparent frequency?

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Explanation

$ By doppler's effect , { f_L \over f_S} = \left( { \nu + \nu_L \over \nu + \nu_S} \right) $ $ \therefore { f_L \over f_S} = { \nu + \nu_L \over \nu } = { \nu + \nu/5 \over \nu } = { 6 \over 5 } $ $ \therefore Fractional increase = { f_L - f_S \over f_S } = { f_L \over f_S } -1 = { 6 \over 5} -1 = {1 \over 5} $ $ \therefore Percentage increase = { 100 \over 5} = 20 \% $

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The speed of sound in Oxygen $( O_2)$ at a certain temperature is 460 m/s. The speed of sound in helium at the same temperature will be…..$ms^{- 1}$ . (Assume both gases to be ideal)

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Explanation

$ \nu = \sqrt { \gamma P \over \rho } = \sqrt { \gamma RT \over M} $ $ \therefore { \nu_2 \over \nu_1 } = \sqrt { { \gamma_{He} \over 4 } \times { 32 \over \gamma_{02} } }$

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In a longitudinal wave, pressure variation and displacement variation are………

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Explanation

In a longitudinal wave, pressure is maximum where displacement is minimum. Therefore pressure and displacement variations are $180 ^\circ $ out of phase

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A tuning fork of frequency 480 hz produces 10 beats/s when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before?

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Explanation

Frequency of tuning fork $f_1= 480 Hz$. Number of beats $s^{-1}$, n = 10 Frequency of string $f_2 = (480 + 10)Hz$. Aslight increase in tension increase $f_2$. $f_2 = 480 - 10 = 470 Hz.$

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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 ……..

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Explanation

As $sin(90 \pm \theta ) = cos \theta $ The phase difference between the two waves is $ \pi /2 $

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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.

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Explanation

$ \nu = \sqrt T = \sqrt { 500 \over 0.2 } = 50 ms^{-1} $

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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

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Explanation

$ 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 } $

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The tension in a wire is decreased by 19%, then the percentage decrease in frequency will be ………

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Explanation

$ { 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 \%$

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