Question

The equation of a simple harmonic wave is given by
$y = 25\pi \sin (25\pi t - \dfrac{\pi }{2}x)$
Where $x$ and $y$ are in meters and $t$ is in seconds the ratio of maximum particle velocity to the wave velocity is:
(A) $\dfrac{3}{2}\pi$
(B) $3\pi$
(C) $\dfrac{{2\pi }}{3}$
(D) $2\pi$

Answer 1

We need to compare the equation of a simple harmonic motion given to the general equation of a simple harmonic motion. Particle velocity can be gotten from the derivative of the simple harmonic motion equation with respect to time.
Formula used: In this solution we will be using the following formulae;
$y = A\sin (\omega t - kx)$ where $y$ is the particle position at any instant in time $t$ and position $x$ along the x axis, $A$ is the amplitude of the wave, this is the general equation of a simple harmonic motion.
${v_p} = \dfrac{{\partial y}}{{\partial t}}$ where ${v_p}$ is the particle velocity, and $\dfrac{{\partial y}}{{\partial t}}$ represents instantaneous change in y displacement of the particle per unit time only even in the presence of other variables.
$v = \dfrac{\omega }{k}$ where $v$ is wave velocity, $\omega$ is the angular frequency, and $k$ is the wave number.

Complete Step-by-Step Solution:
From the given equation $y = 3\sin \dfrac{\pi }{2}(50t - x)$, it can be rewritten as
$y = 3\sin (25\pi t - \dfrac{\pi }{2}x)$
We are asked to find the ratio of the maximum particle velocity to the wave velocity.
First, we compare the general equation of the simple harmonic motion, which can be given as
$y = A\sin (\omega t - kx)$ where $y$ is the particle position at any instant in time $t$ and position $x$ along the x axis.
We see that, $\omega = 25\pi$ and $k = \dfrac{\pi }{2}$
The wave velocity can be given as
$v = \dfrac{\omega }{k}$
Hence,
$v = \dfrac{{25\pi }}{{\dfrac{\pi }{2}}} = 50m/s$
Now, particle velocity can be given by
${v_p} = \dfrac{{\partial y}}{{\partial t}}$
Hence, differentiating the given equation with respect to time, we have,
$\dfrac{{\partial y}}{{\partial t}} = 3 \times 25\pi \cos (25\pi t - \dfrac{\pi }{2}x)$
$\Rightarrow \dfrac{{\partial y}}{{\partial t}} = 75\pi \cos (25\pi t - \dfrac{\pi }{2}x)$
Hence, the maximum particle velocity is ${v_{p\max }} = 75\pi$
Then, the ratio of maximum particle velocity to wave velocity is
$\dfrac{{{v_{p\max }}}}{v} = \dfrac{{75\pi }}{{50}} = \dfrac{3}{2}\pi$

Hence, the correct option is A

Note: for clarity, the ${v_{p\max }} = 75\pi$ can be shown from the observation that $\cos (25\pi t - \dfrac{\pi }{2}x)$ will either be equal to one or less than 1. Hence, the maximum must be when $\cos (25\pi t - \dfrac{\pi }{2}x) = 1$.