Answer

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**Hint:**The pressure amplitude of a wave is equal to the product of the bulk modulus of the fluid, displacement amplitude and the wave number of the wave. Recall that the velocity can be given as the square root of the ratio of the bulk modulus to the density of the medium.

Formula used: In this solution we will be using the following formulae;

\[v = \sqrt {\dfrac{B}{\rho }} \] where \[v\] is the speed of sound in a fluid, \[B\] is the bulk modulus of the fluid, and \[\rho \] is the density.

\[{P_0} = BAk\] where \[P\] is the pressure amplitude of a sound wave, \[A\] is the displacement amplitude, \[k\] is the wave number of the wave.

\[k = \dfrac{{2\pi }}{\lambda }\] where \[\lambda \] is the wavelength of the wave.

\[v = f\lambda \] where \[v\] is the speed of the wave, \[f\] is the frequency.

**Complete Step-by-Step Solution:**

Given the pressure amplitude, we are asked to find the displacement amplitude. Generally, the formula of the pressure amplitude is given as

\[{P_0} = BAk\] where \[B\] is the bulk modulus of the fluid the wave travels, \[A\] is the displacement amplitude, \[k\] is the wave number of the wave.

To calculate \[B\], we recall that

\[v = \sqrt {\dfrac{B}{\rho }} \]

\[ \Rightarrow B = {v^2}\rho \]

Hence, by inserting known values, we have

\[B = {340^2} \times 1.29 = 149124Pa\]

To calculate \[k\], we recall that

\[k = \dfrac{{2\pi }}{\lambda }\] but \[v = f\lambda \], where \[v\] is the speed of the wave, \[f\] is the frequency and \[\lambda \] is the wavelength of the wave, hence,

\[k = \dfrac{{2\pi f}}{v} = \dfrac{{2\pi \left( {100} \right)}}{{340}} = 1.8480{m^{ - 1}}\]

Hence, from \[{P_0} = BAk\]

\[A = \dfrac{{{P_0}}}{{Bk}}\]

\[ \Rightarrow A = \dfrac{{10}}{{149124 \times 1.84800}} = 3.63 \times {10^{ - 5}}m\]

**Hence, the correct answer is A**

**Note:**Alternatively, to avoid time consuming intermediate calculations and approximation errors, we used find the final expression before inserting all known values, as in;

From

\[A = \dfrac{{{P_0}}}{{Bk}}\]

Considering that \[B = {v^2}\rho \] and \[k = \dfrac{{2\pi f}}{v}\] we can substitute into the equation above. Hence,

\[A = \dfrac{{{P_0}}}{{\left( {{v^2}\rho } \right)\left( {\dfrac{{2\pi f}}{v}} \right)}}\]

Simplifying by cancelling \[v\], we have

\[A = \dfrac{{{P_0}}}{{2\pi fv\rho }}\]

Hence, by inserting values, we get

\[ \Rightarrow A = \dfrac{{10}}{{2\pi \times 100 \times 340 \times 1.29}} = 3.63 \times {10^{ - 5}}m\]

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