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.

Answer 1

Ratio of displacement amplitude and wavelength of the wave is $1.7 \times 10^-^5$ m

Explanation:

Given, Here $r_y = 6.0 \times 10^-^5$ m

a) Given $\frac {2\pi}{\lambda}$= 1.8

or $\lambda = \frac {2\pi}{1.8}$

So, $\frac {r_y}{\lambda} = \frac {6.0 \times 1.8 \times 10^-^5 m/s}{2\pi}$

= $1.7 \times 10^-^5$ m

b) Let velocity amplitude = $V_y$

V = $\frac {dy}{dt} = 3600\ cos (600t - 1.8) \times 10^-^5$ m/s  

Here, $V_y = 3600 \times 10^-^5$ m/s

Again $\lambda = \frac {2\pi}{1.8}$ and  

$T = \frac {2\pi}{600}$ or Wave Speed

= v = $\frac {\lambda}{T} = \frac {600}{1.8}$

= $\frac {1000}{3}$m/s

So ratio of $(\frac {V_y}{v}) = \frac {3600 \times 3 \times 10^-^5}{1000}$