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Question: A sound wave of wavelength \(40\,cm\) travels in air. If the difference between the maximum and mini...

A sound wave of wavelength 40 cm40\,cm travels in air. If the difference between the maximum and minimum pressures at a given point is 1×10−3 Nm−21 \times {10^{ - 3}}\,N{m^{ - 2}} . What will be the amplitude of vibration of the particle of the medium, the bulk modulus of air is 1.4×10−10 Nm−21.4 \times {10^{ - 10}}\,N{m^{ - 2}} ?
(A) 4.4×10−10 m4.4 \times {10^{ - 10}}\,m
(B) 3.3×10−10  m3.3 \times {10^{ - 10\,}}\,m
(C) 1.1×10−10  m1.1 \times {10^{ - 10\,}}\,m
(D) 2.2×10−10  m2.2 \times {10^{ - 10\,}}\,m

Explanation

Solution

Hint From the given details, use the formula of the pressure amplitude and get that value. Substitute the obtained pressure amplitude, calculated wave number and other given data in the amplitude formula to know the value of the amplitude.
Useful formula
(1) The formula of the pressure amplitude is given by
p=P2p = \dfrac{P}{2}
Where pp is the pressure amplitude, PP is the difference in the maximum and the minimum pressure of the given point.
(2) The formula of the amplitude is given by
s=pBks = \dfrac{p}{{Bk}}
Where ss is the amplitude of the vibration, BB is the bulk modulus of the air and kk is the wave number.
(3) The wave number is given by
k=2πλk = \dfrac{{2\pi }}{\lambda }
Where λ\lambda is the wavelength of the sound in air.

Complete step by step solution
The wavelength of the sound wave in air, λ=40 cm=40×10−2 m\lambda = 40\,cm = 40 \times {10^{ - 2}}\,m
Difference in the pressure at a point is, P=Pmax⁡−Pmin⁡=1×10−3 Nm−2P = {P_{\max }} - {P_{\min }} = 1 \times {10^{ - 3}}\,N{m^{ - 2}}
The bulk modulus of the air, B=1.4×10−10 Nm−2B = 1.4 \times {10^{ - 10}}\,N{m^{ - 2}}
Using the formula of the pressure amplitude,
p=P2p = \dfrac{P}{2}
Substituting the known values,
p=1×10−5 Nm−22p = \dfrac{{1 \times {{10}^{ - 5}}\,N{m^{ - 2}}}}{2}
p=5×10−2 Nm−2p = 5 \times {10^{ - 2}}\,N{m^{ - 2}}
Substitute the value of the pressure amplitude in the amplitude formula,
s=pBks = \dfrac{p}{{Bk}}
Substituting known values,
s=5×10−2 1.4×105ks = \dfrac{{5 \times {{10}^{ - 2}}\,}}{{1.4 \times {{10}^5}k}} ---------------(1)
Using the formula of wave number, and substitute the wavelength.
k=2π40×10−2k = \dfrac{{2\pi }}{{40 \times {{10}^{ - 2}}}} --------------(2)
Substitute the equation (2) in (1)
s=5×10−2 ×40×10−21.4×105×2πs = \dfrac{{5 \times {{10}^{ - 2}}\, \times 40 \times {{10}^{ - 2}}}}{{1.4 \times {{10}^5} \times 2\pi }}
By simplifying the above step, we get
s=2.2×10−10 ms = 2.2 \times {10^{ - 10}}\,m

Thus the option (D) is correct.

Note The sound waves can transmit through air, water or other solid particles. It is a little shocking that the sound travels faster and easier in the water than in air. This is because the particles of the water are densely arranged when compared to the air, hence the vibration of the sound wave can be transmitted faster.