Question

If the maximum speed of a particle in a medium carrying a traveling wave is $V_0$, then find the speed of the particle when displacement is half of maximum value.

Answer 1

Hint: Here, as we know that displacement can be written as $y = A\sin \left( {\omega t - kx} \right)$, then on differentiating this equation we can find the value of the velocity i.e. $V = \omega A\cos \left( {\omega t - kx} \right)$, after that we can easily find the value of maximum velocity i.e. ${V_0} = A\omega $, now given that $y = \dfrac{A}{2}$, substituting these values we will get the desired result. 

 
Complete step by step answer

As, we know that the displacement of travelling wave can be written as

$y = A\sin \left( {\omega t - kx} \right)$ ……………… (1)

Where, A is the amplitude

And ω is the angular velocity

In order to find out the velocity of the travelling wave we can differentiate the equation (1), we get

$ \Rightarrow \dfrac{{dy}}{{dt}} = A\omega \cos \left( {\omega t - kx} \right)$

As the rate of change of displacement with respect to time is the velocity, then above equation can be written as

$ \Rightarrow V = \omega A\cos \left( {\omega t - kx} \right)$ …………………. (2)

As, it is given that maximum velocity of the traveling wave is ${V_0}$ and the value of velocity V will be maximum when $\cos \left( {\omega t - kx} \right)$ is maximum i.e. $\cos \left( {\omega t - kx} \right) = 1$

Substitute the values in equation (2), we get

$ \Rightarrow {V_0} = A\omega $ ………………………… (3)

Now, we have to find the speed of the particle when displacement is half of the maximum value i.e.

$y = \dfrac{A}{2}$, A is the amplitude or maximum displacement of the traveling wave

Now substitute the value of y in equation (1), we get

$\Rightarrow \dfrac{A}{2} = A\sin \left( {\omega t - kx} \right) $

$\Rightarrow \sin \left( {\omega t - kx} \right) = \dfrac{1}{2} $

As, we also know that $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ then from the above equation, we can write

$ \Rightarrow \omega t - kx = \dfrac{\pi }{6}$

Put this value in equation (2), we get

$ \Rightarrow V = A\omega \cos \dfrac{\pi }{6}$

As we also know that $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$, and using the equation (3), therefore above equation will become

$ \Rightarrow V = \dfrac{{\sqrt 3 }}{2}{V_0}$

This is the required speed of the particle.

Note
In this question, it should be notice that when the quantity is given in the form of trigonometry terms then that quantity will only be maximum when that trigonometry term has maximum value and for that case the value of whole trigonometry term should be maximum it is not necessary that the angle of that term is maximum, for example in above equation the value of V will only be maximum when the value of whole trigonometry term i.e. $\cos \left( {\omega t - kx} \right)$ is maximum i.e. $\cos \left( {\omega t - kx} \right) = 1$ for this value the angle is zero degrees instead of $90^\circ$.