Question

If the speed of the wave shown in the figure is 330 m/s in the given medium, then the equation of the wave propagation in the positive x-direction will be ................
(a) y = 0.05 sin 2π (4000t - 12.5x) m
(b) y = 0.05 sin 2π (4000t - 122.5x) m
(c) y = 0.05 sin 2π (3300t - 10x) m
(d) y = 0.05 sin 2π (3300t - 10x) m

Answer 1

Answer: $y = 0.05\sin 2\pi (3300t-10x)\ m$

Explanation:

Given that,

Speed of the wave c = 330 m/s

The general equation of wave propagation is

$y = a\sin2\pi(\omega t- kx)$

According to graph of wave propagation

Amplitude a = 0.05

The wave length is

$2.5\lambda = 0.25$

$\lambda = 0.1 m$

Now, the value of $\omega$ is

$\omega = 2\pi\dfrac{c}{\lambda}$

$\omega = 2\pi(3300)$

Now, the value of k is

$k = \dfrac{2\pi}{0.1}$

$k = 2\pi 10$

Hence, the equation of the wave propagation in the positive x- direction will be $y = 0.05 \sin 2\pi (3300t-10x)\ m$

Answer 2

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