Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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 - Physics

Sum

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.

#### 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.

$\text { Velocity v }= \frac{dy}{dt}$

$v = \frac{d\left[ 6 \sin \left( 600 t - 1 . 8 x \right) \right]}{dt}$

$\Rightarrow v = 3600 \cos (600t - 1 . 8x) \times {10}^{- 5} m/s$

$\text { Amplitute } V_y = 3600 \times {10}^{- 5} m/s$

$\text { Wavelength: }$

$\lambda = \frac{2\pi}{1 . 8}$

$\text { Time period: }$

$T = \frac{2\pi}{\omega}$

$\Rightarrow T = \frac{2\pi}{600}$

$\text { Wave speed v } = \frac{\lambda}{T}$

$\Rightarrow v = \frac{600}{1 . 8} = \frac{100}{3} m/s$

$\text { Required ratio: }$

$\left( \frac{V_y}{v} \right) = \frac{3600 \times 3 \times {10}^{- 5}}{1000} = 1 . 1 \times {10}^{- 4} m$

Concept: Wave Motion
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 8 | Page 353