Question

The transverse displacement of a particle at $x=0.360m$ at time $t=0.15s$ is $-510\times {10}^{-x}m$. Find $x$.

Answer 1

A left travelling transverse wave given by $y=Asin(kx+\omega t)$

where wave number,$k=\frac{2\pi }{\lambda }=\frac{2\pi }{0.32}rad/m$

Speed of wave=$\lambda \nu =8m/s$

$⟹\nu =\frac{8}{0.32}Hz=25Hz$

$⟹\omega =2\pi \nu =\frac{2\pi }{0.04s}$

Thus $y=\left(0.07m\right)sin2\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$

Thus $y(0.36m,0.15s)=\left(0.07m\right)sin2\pi (\frac{0.36m}{0.32m}+\frac{0.15s}{0.04s})$

$=-0.0510m$