Question

The time must elapse from the instant $0.15s$ until the particle at $x=0.360m$ next has maximum upward displacement is $5\times {10}^{-x}sec$. Find the value of $x$.

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=$\lambda \nu =8m/s$

$⟹\nu =\frac{8}{0.32}Hz=25Hz$

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

We see that the upward displacement maxima occurs for integrals multiples of $2\pi$.

Thus $n=\frac{0.36}{0.32}+\frac{t}{0.04}$

where n is an integer.

For t=0.15s, $n=4.875$

Closest next integer value is $n=5$

Thus $t(n=5)=0.155s$

Thus time until it happens=$0.155s-0.15s=0.005s$