Question

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

Answer 1

As the displacement of wave is given by,
$y=A\sin (\omega t-kx)$
$\because k=2\pi /\lambda$ and $\omega =2\pi /T$
$\frac{y}{A}=\sin 2\pi (\frac{t}{T}-\frac{x_1}{\lambda })$
For the first particle displacement equation can be written as:
$y_1=1,A=2,$
$1=2\sin 2\pi (\frac{t}{T}-\frac{x_1}{\lambda })$
$2\pi (\frac{t}{T}-\frac{x_1}{\lambda })={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}rad\dots \left(i\right)$
For the second particle displacement equation can be written as:
$y_2=\sqrt{2,}A=2,$
$\sqrt{2}=2\sin 2\pi (\frac{t}{T}-\frac{x_2}{\lambda })$
$2\pi (\frac{t}{T}-\frac{x_2}{\lambda })={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}rad\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
$\frac{t}{T}-\frac{x_1}{\lambda }=\frac{1}{12}$
And $\frac{t}{T}-\frac{x_2}{\lambda }=\frac{1}{8}$
Subtract, $\frac{x_1}{\lambda }-\frac{x_2}{\lambda }=\frac{1}{24}$
$(x_1-x_2)=\frac{\lambda }{24}$
Final answer: $\left(24\right)$