Question

A progressive wave is represented by $x = A\sin \omega t$ . If $v$ and $a$ represent the velocity and acceleration of the wave, then obtain a relation between velocity and acceleration of the wave.
A) $\dfrac{a}{v} = - \tan \omega t$
B) $\dfrac{a}{v} = - \omega \tan \omega t$
C) $\dfrac{v}{a} = - \tan \omega t$
D) $\dfrac{v}{a} = - \omega \tan \omega t$

Answer 1

Hint: Velocity is the rate of change of displacement and acceleration is the rate of change of velocity. Therefore, the first derivative of the displacement, $x = A\sin \omega t$, concerning time gives the velocity of the wave. Differentiating velocity with respect to time gives us the acceleration of the wave.

Complete step by step solution:
Step 1: Obtain an expression for the velocity of the wave.
The displacement of the progressive wave is expressed as $x = A\sin \omega t$ where $\omega $ is angular velocity and $A$ is the amplitude of the wave and it is constant.
Since velocity is the rate of change of displacement, we take the first derivative of the displacement i.e., $\dfrac{d}{{dt}}\left( x \right)$ .
The first derivative of $x$ is given by, $\dfrac{d}{{dt}}\left( x \right) = \dfrac{d}{{dt}}\left( {A\sin \omega t} \right)$ .
Here, the amplitude $A$ of the wave is constant in time.
Then, $\dfrac{d}{{dt}}\left( {A\sin \omega t} \right) = A\dfrac{d}{{dt}}\left( {\sin \omega t} \right) = A\omega \cos \omega t$ .
Thus the velocity of the wave is $v = A\omega \cos \omega t$.
Step 2: Obtain an expression for the acceleration of the wave.
The velocity of the wave is obtained as $v = A\omega \cos \omega t$ . Since acceleration is the rate of change of velocity, we take the first derivative of the velocity i.e., $\dfrac{d}{{dt}}\left( v \right)$ .
The first derivative of $v$ is given by, $\dfrac{d}{{dt}}\left( v \right) = \dfrac{d}{{dt}}\left( {A\omega \cos \omega t} \right)$ .
Here, the amplitude $A$ of the wave is constant in time.
Then, $\dfrac{d}{{dt}}\left( {A\omega \cos \omega t} \right) = A\dfrac{d}{{dt}}\left( {\cos \omega t} \right) = - A{\omega ^2}\sin \omega t$ .
Thus the acceleration of the wave is $a = - A{\omega ^2}\sin \omega t$ .
Step 3: Find the relation between velocity and acceleration.
We have velocity $v = A\omega \cos \omega t$ and acceleration $a = - A{\omega ^2}\sin \omega t$ .
Express the ratio of acceleration over velocity to get, $\dfrac{a}{v} = \dfrac{{ - A{\omega ^2}\sin \omega t}}{{A\omega \cos \omega t}}$ .
Cancel out the similar terms to get, $\dfrac{a}{v} = \dfrac{{ - \omega \sin \omega t}}{{\cos \omega t}} = - \omega \tan \omega t$ .

$\therefore $ The relation between velocity $v$ and acceleration $a$ is $\dfrac{a}{v} = - \omega \tan \omega t$.

Note:
Acceleration of the wave is obtained by taking the first derivative of the velocity with respect to time i.e., $\dfrac{d}{{dt}}\left( v \right)$. This is same as taking the second derivative of the displacement with respect to time i.e., $\dfrac{{{d^2}}}{{d{t^2}}}\left( x \right)$ because $\dfrac{{{d^2}}}{{d{t^2}}}\left( x \right) = \dfrac{d}{{dt}}\left( {\dfrac{d}{{dt}}\left( x \right)} \right)$ and $\dfrac{d}{{dt}}\left( x \right) = v$ .
The derivative of $\sin bt$ with respect to $t$ , where $b$ is the coefficient of $t$ , is given by, $\dfrac{d}{{dt}}\left( {\sin bt} \right) = \left( {\cos bt} \right) \times b$ .
The derivative of $\cos bt$ with respect to $t$ , where $b$ is the coefficient of $t$ , is given by, $\dfrac{d}{{dt}}\left( {\cos bt} \right) = - \left( {\sin bt} \right) \times b$ .