Question

Transverse waves on a string have wave speed $12.0$ m/s, amplitude $0.05m$ and wavelength $0.4m$. The waves travel in the $+x$ direction and at $t=0$, the $x=0$ end of the string has zero displacement and is moving upwards. The time must elapse from the instant 0.15 s until the point at $x=0.25m$ has zero displacement is $4.2\times {10}^{-x}sec$. Find $x$.

Answer 1

A wave traveling in +x direction is represented by $y=Asin(\omega t-kx)$

where $\omega =2\pi \nu$

and $k=\frac{2\pi }{\lambda }$

We know that speed of wave=$v=\lambda \nu$

Thus here

$\omega =2\pi \times \frac{v}{\lambda }=2\pi \times \frac{12}{0.4}=60\pi s^{-1}$

and $k=\frac{2\pi }{0.4}=5\pi m^{-1}$

Thus wave is $y=\left(0.05m\right)sin\left(\right(60\pi s^{-1})t-(5\pi m^{-1}\left)x\right)$

At the given moment, displacement of point at $x=0.25m$ is zero.

Thus $0=0.05sin\left(\right(\omega t-kx\left)\right)$

$⟹\omega t-kx=n\pi$

Put $x=0.25m$

We get $t=0.1542s$ for $n=8$

Thus the displacement of the point is zero after $0.1542s-0.15s=4.2ms$ for the first time after the instant.