Question

Two small objects are $15\text{m}$ apart on the water surface of a lake. Each object pops up and down with a period of $4.5\text{second}$ due to wave motion on water surface.
When one object is at its highest point, the another one is at its lowest point. Consider that objects are always within a single cycle of wave, the speed of wave is
(Assume transverse surface waves)

• A. $2.5\text{m/s}$
• B. $\frac{10}{3}\text{m/s}$
• C. $12\text{m/s}$
• D. $\frac{20}{3}\text{m/s}$

Answer 1

The correct option is D $\frac{20}{3}\text{m/s}$

Assuming transverse displacement of objects.
Both the objects are having a phase difference of $\pi$, because while one is at its maximum (highest) point the other one is at lowest position.

Seperation between particles = $\frac{\lambda }{2}$
$\Rightarrow 15=\frac{\lambda }{2}\text{or}\lambda =30\text{m}$
and Time period, $T=4.5\text{sec}$
$\Rightarrow v=f\lambda =\frac{\lambda }{T}$
$\therefore v=\frac{30}{4.5}=\frac{20}{3}\text{m/s}$ (wave speed)