Question

A wave of frequency $400Hz$ has a velocity of $320m{s}^{-1}$. The distance between the particles differing in phase by ${90}^{\circ }$ is

• A. $80cm$
• B. $60cm$
• C. $40cm$
• D. $20cm$

Answer 1

The correct option is D $20cm$

Let,
Frequency of wave is, $f=400Hz$
Velocity of wave is, $v=320m{s}^{-1}$
The phase difference is, $\varphi ={90}^{\circ }=\frac{\pi }{2}$
The wavelength of wave is, $\lambda$
$\lambda =\frac{v}{f}$
$\lambda =\frac{320}{400}$
$\lambda =0.8m$
Now, the phase difference between two particles in a wave is
$\frac{\varphi }{2\pi }=\frac{x}{\lambda }$
where, x is the distance between two particles in a wave.
$x=\frac{\lambda \varphi }{2\pi }$
$x=\frac{0.8\left(\frac{\pi }{2}\right)}{2\pi }$
$x=\frac{0.8}{4}=0.2m$
$x=20cm$