Physics - Wave on String Question with Solution | TestHub

The given wave equation is $y=60\sin (1200t-6x)$.

Comparing this with the standard wave equation $y=A\sin (\omega t-kx)$, we get:
Amplitude $A=60\text{ mum}=60\times 10^{-6}\text{ m}$
Angular frequency $\omega =1200\text{ rad/s}$
Wave number $k=6{\text{ m}}^{-1}$

The maximum particle velocity is given by $v_{p,\text{max}}=A\omega$.
The wave velocity is given by $v_w=\frac{\omega }{k}$.

The ratio of maximum particle velocity to wave velocity is:
$\text{Ratio}=\frac{v_{p,\text{max}}}{v_w}=\frac{A\omega }{\frac{\omega }{k}}=Ak$

Substituting the values of $A$ and $k$:
$\text{Ratio}=60\times 10^{-6}\times 6=360\times 10^{-6}=3.6\times 10^{-4}$

Answer: $3.6\times 10^{-4}$