Question

The maximum particle velocity is $3$ times the wave velocity of a progressive wave. If the amplitude of the particle is $a$, the phase difference between the two particles separated by a distance of ${}^{\prime }{x}^{\prime }$ is

• A. $\frac{x}{a}$
• B. $\frac{3x}{a}$
• C. $\frac{3a}{x}$
• D. $\frac{3\pi x}{a}$

Answer 1

The correct option is B $\frac{3x}{a}$

Given:
amplitude $=a;\&\Delta x=x$

Let,
$v_{p,m}=$ maximum velocity of particle
$v_w=$ wave velocity

For a progressive wave, we know that

$v_{p,m}=A\omega =a\times 2\pi f$ and $v_w=f\lambda$

From question,

$v_p=3v_w$

$\Rightarrow a\times 2\pi f=3f\lambda$

$\Rightarrow \lambda =\frac{2\pi a}{3}$

So, phase difference $\left(\Delta \varphi \right)$ for $\Delta x$ path difference is

$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$

$\Delta \varphi =\frac{2\pi }{\frac{2\pi a}{3}}\times x=\frac{3x}{a}$

Hence, option $\left(b\right)$ is the correct answer.