Question

In a transverse wave the distance between a crest and through at the same place is $1.0{\rm{ cm}}$. The next crest appears at the same place after the interval of $0.4\;{\rm{s}}$. The maximum speed of the vibrating particles in the medium is:
A) $\dfrac{{3\pi }}{2}{\rm{cm/s}}$
B) $\dfrac{{5\pi }}{2}{\rm{cm/s}}$
C) $\dfrac{\pi }{2}{\rm{cm/s}}$
D) $2\pi \;{\rm{cm/s}}$

Answer 1

The maximum speed of the vibrating particles can be calculated with the help of the general equation of a wave. If we differentiate the general equation of a wave, we can get the general equation of velocity.

Complete step by step solution:
The distance between the crest and trough is $1.0{\rm{ cm}}$ which means that the amplitude is half of $1.0{\rm{ cm}}$. This can evaluate the value of amplitude as given below,
$A$ = \dfrac{{1{\rm{ cm}}}}{2}\\\
$A$ = $0.5\;{\rm{cm}}$
The interval at which the next crest appears is nothing but the time period, therefore, the time period of the given wave is $0.4\;{\rm{s}}$.
We can calculate the value of angular frequency with the help of time period.
$\omega = \dfrac{{2\pi }}{T}$
We will now substitute the known values in the above equation of angular frequency.
\omega = \dfrac{{2\pi }}{{0.4\;{\rm{s}}}}\\\
$\Rightarrow$ $5\pi \;{\rm{rad/s}}$
Here, the time interval is $T$.
We know that the general equation of a wave is given as $y = A\sin( \omega t + kx)$.
So, for maximum velocity, we will differentiate the above equation with respect to time.
$v$ = \dfrac{{dy}}{{dt}}\\\
$\Rightarrow$ \dfrac{{d\left( {A\sin \omega t + kx} \right)}}{{dt}}\\\
$\Rightarrow$ $A\omega \left( {\cos \omega t + kx} \right)$
The maximum value of the equation is ${v_{\max }} = A\omega$.
The equation of maximum velocity is evaluated and now we can substitute the values to get maximum velocity.
{v_{\max }} = 0.5 \times 5\pi \\\
$\Rightarrow$ $\dfrac{{5\pi }}{2}\;{\rm{cm/s}}$
Thus, the maximum speed of the vibrating particles in the given medium is calculated to be $\dfrac{{5\pi }}{2}\;{\rm{cm/s}}$.

Thus, From the given options, only option B is correct.

Note: The step in which the equation for maximum velocity comes to be $A\omega$ is a tricky method. We should remember that the maximum value of any sine or cosine function is 1, so in order to get maximum value of the equation $A\omega \left( {\cos \omega t + kx} \right)$, the cosine function is taken as 1.