Question

98. The maximum pressure variation that the human ear
can tolerate in loud sound is about 30 Nm-2. The
corresponding maximum displacement for a sound
wave in air having a frequency of 10 Hz is (take
velocity of sound in air as 300 ms -- and density of
air 1.5 kg m-3)
​

Answer 1

Answer:

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Answer 2

Answer: Option 4)  is correct.

Explanation:

(\Delta P_{max}) = BAK \Rightarrow A =\frac{\Delta P}{BK}

v = \frac{\omega}{k}

k = \omega \sqrt{\frac{\rho}{B}} \Rightarrow A = \frac{\Delta P_{max}}{2\pi f \rho v } = \frac{10^{-4}}{3 \pi} m

 

Equation of sound wave -

\Delta P= \Delta P_{max}\cdot \sin\left [ \omega \left ( t-\frac{x}{V} \right ) \right ]

 - wherein

{\Delta P= variation in pressure at a point

\Delta P_{max}= maxmium variation in pressure }

 

 

Option 1)

\frac{2\pi}{3} \times 10^{-2}\; m

This is incorrect.

Option 2)

\frac{2\times 10^{-4}}{\pi}\; m

This is incorrect.

Option 3)

\frac{\pi}{3} \times 10^{-2}\; m

This is incorrect.

Option 4)

\frac{10^{-4}}{3\pi}\; m

This is correct.