Question

A simple harmonic wave of amplitude 2 units is travelling along x - axis. At a given instant, displacements for two different particles on the wave are 1 unit and $\sqrt{2}$
units. If the wavelength of wave is $\lambda$, find the distance between them.

• A. λ/12
• B. λ/8
• C. λ/24
• D. λ/16

Answer 1

The correct option is C

λ/24

$\frac{y}{A}$ = sin 2 $\pi$ ($\frac{t}{T}$ - $\frac{x}{\lambda }$)

$y_1$ = 1 , A = 2 , 2 $\pi$ ($\frac{t}{T}$ - $\frac{x_1}{\lambda }$) = ${sin}^{-1}$ ($\frac{1}{2}$) = $\frac{\pi }{6}$ rad.

$y_1$ = $\sqrt{2}$ , A = 2 , 2 $\pi$ ($\frac{t}{T}$ - $\frac{x_2}{\lambda }$) = ${sin}^{-1}$ ($\frac{1}{\sqrt{2}}$) = $\frac{\pi }{4}$ rad.

$\frac{t}{T}$ - $\frac{x_1}{\lambda }$ = $\frac{1}{12}$ ; $\frac{t}{T}$ - $\frac{x_2}{\lambda }$ = $\frac{1}{8}$

Subtract , $\frac{x_1}{\lambda }$ - $\frac{x_2}{\lambda }$ = $\frac{1}{24}$ ; ($x_1$ - $x_2$) = $\frac{\lambda }{24}$