Question

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

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

Answer 1

The correct option is C $y(x,t)=\left(0.02m\right)\sin [\left(2\pi m^{-1}\right)x-\left(10\pi s^{-1}\right)t]m$

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