Question

A sinusoidal wave travelling in the positive direction on stretched string has amplitude $20 \,cm$, wavelength $1\, m$ and wave velocity $5\, m / s$. At $x=0$ and $t=0$, it is given that $y=0$ and $\frac{d y}{d t}< 0 .$ Find the wave function $y(x, t)$.

• A. $y(x, t)=(0.02 m ) \sin \left[\left(2 \pi m ^{-1}\right) x+\left(10 \pi s ^{-1}\right) t\right] m$
• B. $y(x, t)=(0.02 m ) \cos \left[\left(10 \pi s ^{-1}\right) t+\left(2 \pi m ^{-1}\right) x\right] m$
• C. $y(x, t)=(0.02 m ) \sin \left[\left(2 \pi m ^{-1}\right) x-\left(10 \pi s ^{-1}\right) t\right] m$
• D. $y(x, t)=(0.02 m ) \sin \left[\left(\pi m ^{-1}\right) x+\left(5 \pi s ^{-1}\right) t\right] m$

Answer 1

Answer: (C)

We start a general form for a rightward moving wave,
$y(x, t)=A \sin (k x-\omega t+\phi)$
The given amplitude is $A=2 \,cm =0.02 \,m$
The wavelength is given as
$\lambda=1 \, m$
Wave number $=k=2 \pi / \lambda=2 \pi m ^{-1}$
Angular frequency,
$\omega=v k=10 \, \pi \, rad / s$
From Eq. (i),
$y(x, t)=(0.02) \sin [2 \pi(x-5 t)+\phi]$
$\because $ For $x=0, t=0$
$y=0$ and $\frac{\partial y}{\partial t}<0$
i.e. $0.02 \sin \phi=0$
(as $y=0$ )
and $-0.2 \pi \cos \phi<0$
From these conditions, we may conclude that
$\phi=2 n \pi$, where $n=0,2,4,6, \ldots$
Therefore,
$y(x, t)=(0.02 \,m ) \sin \left[\left(2 \,\pi \, m ^{-1}\right) x-\left(10 \, \pi \, s ^{-1}\right) t\right] m$