Question

A transverse wave is described by the equation $y = {y_ \circ }\sin 2\pi \left( {ft - \dfrac{x}{t}} \right)$. The maximum velocity of the particle is equal to four times the wave velocity, if
A) $\lambda = \dfrac{{\pi {y_ \circ }}}{4}$
B) $\lambda = \dfrac{{\pi {y_ \circ }}}{2}$
C) $\lambda = \pi {y_ \circ }$
D) $\lambda = \dfrac{{2\pi }}{{{y_ \circ }}}$

Answer 1

Hint: This question can be approached by the mindset that what transverse waves are, and what their mathematical equations are. Additionally, students need to have a firm grip on concepts like wave velocity, particle velocity and their associated relation with each other in mathematical equations to solve this problem.

Formula Used:
${\text{wave velocity, }}v = \dfrac{{{\text{coefficient of t}}}}{{{\text{coefficient of x}}}}$
Maximum particle velocity, ${v_{\max }} = \omega A$

Complete step by step answer:
A transverse wave is defined as a moving wave whose oscillations are perpendicular to the direction of propagation. A simple example of transverse would be the waves created on a horizontal length of string by anchoring one end and moving the other end up and down.
Given equation of transverse wave is:
$y = {y_ \circ }\sin 2\pi \left( {ft - \dfrac{x}{t}} \right)$
The velocity of a wave is mathematically equal to the product of its wavelength and frequency i.e., number of vibrations per second and is independent of the intensity of the wave itself. Also mathematically, wave velocity is given by:
\eqalign{ & {\text{wave velocity, }}v = \dfrac{{{\text{coefficient of t}}}}{{{\text{coefficient of x}}}} \cr & \Rightarrow v = \dfrac{{2\pi f}}{{\dfrac{{2\pi }}{\lambda }}} \cr & \Rightarrow v = \lambda f \cr}
Particle velocity is defined as the velocity of a particle in a medium as it transmits a wave through the medium. The SI unit of particle velocity is the metre per second (m/s).
Mathematically, maximum value of particle velocity is given by:
\eqalign{ & {v_{\max }} = \omega A \cr & \Rightarrow {v_{\max }} = 2\pi f{y_ \circ } \cr}
It is given in the question that ${v_{\max }} = 4v$
Substituting its value in the above equation, we get:
\eqalign{ & 2\pi f{y_ \circ } = 4v \cr & \Rightarrow 2\pi f{y_ \circ } = 4\lambda f \cr & \Rightarrow \lambda = \dfrac{{2\pi f{y_ \circ }}}{{4f}} \cr & \Rightarrow \lambda = \dfrac{{\pi {y_ \circ }}}{2} \cr}
Therefore, the correct answer is B. i.e. $\lambda = \dfrac{{\pi {y_ \circ }}}{2}$

Note: Another approach to solve this equation is to differentiate the given equation of the transverse wave with respect to time. And find the maximum value of particle velocity from this. Then compare it with the given value. Thus solve for the required value of wavelength.