Question

If a wave travelling in positive x-direction with $A=0.2m/s,velocity=360m/s$ and $\lambda =60m$, then correct expression for the wave is:

• A. $y=0.2sin\left[2\pi (6t+\frac{x}{60})\right]$
• B. $y=0.2sin\left[\pi (6t+\frac{x}{60})\right]$
• C. $y=0.2sin\left[2\pi (6t-\frac{x}{60})\right]$
• D. $y=0.2sin\left[\pi (6t-\frac{x}{60})\right]$

Answer 1

The correct option is C $y=0.2sin\left[2\pi (6t-\frac{x}{60})\right]$

General equation for a plane progressive wave travelling along positive x-direction is given b
$y=A\sin (wt-kx)$
$=A\sin \left[2\pi (v-\frac{x}{\lambda })\right]....\left(i\right)$
Here, it is given
$v=360m/s,\lambda =60m$
We know that, frequency $v=\frac{v}{\lambda }=\frac{360}{60}=6Hz$.
Substituting $A=0.2m/s,v=6Hz$
$\lambda =60m$ in Eq. (i), we get
$y=0.2sin\left[2\pi (6t-\frac{x}{60})\right]$.