Question

A sinusoidal wave travelling in the positive direction on stretched string has amplitude 2.0 cm, wavelength 1.0 m and wave velocity 5.0 m/s. At x = 0 and t = 0 it is given that y = 0 and $\frac{\delta y}{\delta t}<0$. Find the wave function y(x, t).

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

Answer 1

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

We start with a general from for a rightward moving
$y(x,t)=Asin(kx-\omega t+\varphi )$
The amplitude given in A = 2.0 cm = 0.02 m.
The wavelength is given as,
$\lambda =1.0m$
Wave number $k=\frac{2\pi }{\lambda }=2\pi m^{-1}$
Angular frequency,
$\omega =vk=10\pi rad/s$
$y(x,t)=\left(0.02\right)sin\left[2\pi \right(x-5.0t)+\varphi ]$
We are told that for x = 0, t = 0,
$y=0and\frac{\delta y}{\delta t}<0$
i.e., $0.02sin\varphi =0$ (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,... ... ... ...
Therefore,
$y(x,t)=\left(0.02m\right)sin\left[\right(2\pi m^{-1})x-(10\pi s^{-1}\left)t\right]m$