Question

A transverse wave is described by the equation $y={{y}_{0}}\sin 2\pi \left( ft-\dfrac{x}{\lambda } \right)$. The maximum particle velocity is equal to four times the wave velocity if:

& \text{A}\text{. }\lambda =\dfrac{\pi {{y}_{0}}}{4} \\\ & \text{B}\text{. }\lambda =\dfrac{\pi {{y}_{0}}}{2} \\\ & \text{C}\text{. }\lambda =\pi {{y}_{0}} \\\ & \text{D}\text{. }\lambda =2\pi {{y}_{0}} \\\ \end{aligned}$$

Answer 1

Hint: Find the wave velocity from the equation and remember two points on the wave always maintain a specific distance. Differentiate the wave equation to find the maximum particle velocity. Compare these two values according to the given condition to find the value of λ.

Formula used:
$y=A\sin (\omega t-kx)$

Complete step by step answer:

We can write a simple wave equation in the form:
$y=A\sin (\omega t-kx)$
Where,
$A$ is the amplitude
$k$ is the propagation constant
$\omega$ is the angular velocity

Two points on the wave always maintain a specific distance. So, it retains its displacement as it moves. So, for a fixed point on the waveform, we must have a constant argument. Hence, we have
$\omega t-kx=$constant

We can differentiate the equation with respect to time,
$\omega -k\dfrac{dx}{dt}=0$
$\Rightarrow \dfrac{dx}{dt}=\dfrac{\omega }{k}$

$\dfrac{\omega }{k}$ gives the wave velocity of the wave.

Hence, in this case the is given by,
$y={{y}_{0}}\sin 2\pi \left( ft-\dfrac{x}{\lambda } \right)$.................(1)

So, we can write that,
$v=\dfrac{f}{\left( \dfrac{1}{\lambda } \right)}=\lambda f$
We can differentiate equation (1),
$v={{y}_{0}}(2\pi f)\cos 2\pi \left( ft-\dfrac{x}{\lambda } \right)$................(2)

Hence, the maximum value of the equation (2) is,
${{v}_{\max }}={{y}_{0}}(2\pi f)$

It is given that,
${{v}_{\max }}=4v$

So, we can write,
$2\pi f{{y}_{0}}=4\lambda f$
$\Rightarrow \lambda =\dfrac{\pi {{y}_{0}}}{2}$
Hence,
$\lambda =\dfrac{\pi {{y}_{0}}}{2}$

So, the correct option is (B).

Note: The quantities wave velocity and particle velocity are different for progressing waves. Wave velocity is the velocity by which the wave progresses through the medium. Wave velocity depends on the properties of the medium. It is given by,
$\dfrac{\omega }{k}$

Whereas, the particle velocity is a changing velocity by which the particles in the medium moves for the successful propagation of the wave. We need to differentiate the given wave equation to find the particle velocity. For example, we can differentiate the conventional form to find the velocity equation.
$v=Aw\cos (\omega t-kx)$

As you can see, both of these quantities are quite different from each other.