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Question
The equation of a travelling sound wave is y = 6.0 sin (600 t − 1.8 x) where y is measured in 10−5 m, t in second and x in metre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of the particles to the wave speed.
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Solution
Given:
Equation of a travelling sound wave is y = 6.0 sin (600 t − 1.8 x),
where y is measured in 10−5 m,
t in second,
x in metre.
Comparing the given equation with the wave equation, we find:
Amplitude A = 6 \[\times\]10-5 m
\[(a) \text{ We have: } \]
\[ \frac{2\pi}{\lambda} = 1 . 8 \]
\[ \Rightarrow \lambda = \left( \frac{2\pi}{1 . 8} \right)\]
\[\text { So, required ratio: } \]
\[ \frac{A}{\lambda} = \frac{6 . 0 \times (1 . 8) \times {10}^{- 5} m/s}{(2\pi)} = 1 . 7 \times {10}^{- 5} m\]
(b) Let Vy be the velocity amplitude of the wave.
\[\text { Velocity v }= \frac{dy}{dt}\]
\[v = \frac{d\left[ 6 \sin \left( 600 t - 1 . 8 x \right) \right]}{dt}\]
\[ \Rightarrow v = 3600 \cos (600t - 1 . 8x) \times {10}^{- 5} m/s\]
\[\text { Amplitute } V_y = 3600 \times {10}^{- 5} m/s\]
\[\text { Wavelength: }\]
\[ \lambda = \frac{2\pi}{1 . 8}\]
\[\text { Time period: } \]
\[T = \frac{2\pi}{\omega}\]
\[ \Rightarrow T = \frac{2\pi}{600}\]
\[\text { Wave speed v } = \frac{\lambda}{T}\]
\[ \Rightarrow v = \frac{600}{1 . 8} = \frac{100}{3} m/s\]
\[\text { Required ratio: } \]
\[\left( \frac{V_y}{v} \right) = \frac{3600 \times 3 \times {10}^{- 5}}{1000} = 1 . 1 \times {10}^{- 4} m\]
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