Question

A sinusoidal wave travelling in the positive direction on a stretched string has an amplitude $20\text{cm}$, wavelength $1.0\text{m}$ and wave velocity $5\text{m/s}$. At $x=0$ and $t=0$ , it is given that $y=0$ and $\frac{\partial y}{\partial t}<0$. Find the wave function $y(x,t)$

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

Answer 1

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

General equation of wave travelling in positive $x-$ direction
$y(x,t)=A\sin (kx-\omega t+\varphi )$
Given that
Amplitude $\left(A\right)=20\text{cm}\text{or}0.02\text{m}$
wavelength $\left(\lambda \right)=1\text{m}$
Angular wave number $\left(k\right)=\frac{2\pi }{\lambda }=2\pi {\text{m}}^{-1}$
Angular frequency $\omega =vk=10\pi \text{rad/s}$
So, $y(x,t)=0.02\sin [2\pi x-10\pi t+\varphi ]........\left(1\right)$

From the data given in the question ,

for $x=0,t=0$ $⟶y=0\&\frac{\partial y}{\partial t}<0$

So, for $y=0$, we get
$0.02\sin \varphi =0\Rightarrow \sin \varphi =n\pi ....\left(2\right)$ Where $n=0,1,2,\mathrm{3.....}$

For , $\frac{\partial y}{\partial t}<0$
$-0.2\pi \cos \varphi <0\Rightarrow 0.2\cos \varphi >0\Rightarrow \cos \varphi >0.......\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$ we get,
$\varphi =2n\pi$ where $n=0,2,4,\mathrm{6....}$
Now we know that
$\sin (2\pi +\theta )=\sin \theta$
Therefore,
$y(x,t)=0.02\sin \left[\right(2\pi x)-(10\pi t\left)\right]\text{m}$