A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (Figure). The waves are of two types: transverse waves traveling at ${\mathit{v}}_{\mathbf{t}}\mathbf{=}\mathbf{50}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and longitudinal waves traveling at ${\mathit{v}}_{\mathbf{t}}\mathbf{=}\mathbf{50}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. If a sudden motion sends out such waves, a scorpion can tell the distance of the beetle from the difference localid="1657274843608" $\mathit{t}$in the arrival times of the waves at its leg nearest the beetle. If localid="1661230422984" $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ms}$, what is the beetle’s distance?

• The speed of the transverse wave,$u=50\mathrm{m}/\mathrm{s}$
• The speed of the longitudinal wave,$v=150\mathrm{m}/\mathrm{s}$
• The time difference between the two signals,$∆t=4.0\mathrm{ms}\mathrm{or}4.0\times {10}^{-3}\mathrm{s}$

The transverse wave is the wave in which the particles of the medium oscillate perpendicular to the direction of propagation of the wave. In a longitudinal wave, the particles of the medium oscillate in the direction that is parallel to the direction of propagation of the wave.

The two types of waves traveling from the beetle to the scorpion move at different speeds. This creates a time difference between the two signals as they reach the scorpion. This time difference in the two signals helps the scorpion locates the prey.

Formula:

The velocity of a wave, $u=\frac{d}{t}$ (i)

The two waves travel the same distance with different speeds. Hence, the time for both transverse and longitudinal waves using equation (i), we get

$\mathrm{Time}\mathrm{for}\mathrm{transverse}\mathrm{wave},t_t=\frac{d}{u}............................\left(1\right)$

$\mathrm{Time}\mathrm{for}\mathrm{longitudinnal}\mathrm{wave},t_t=\frac{d}{v}..........................\left(2\right)$

Hence, using the same equation (i), (1) & (2) and the given values of time difference and velocities, the distance of the beetle will be as given:

$$\begin{gathered} ∆t=\frac{d}{u}-\frac{d}{v} \\ 4\times {10}^{-3}=\frac{d}{50}-\frac{d}{150} \\ 4\times {10}^{-3}=\frac{100d}{150} \\ d=\frac{0.6}{100} \\ =6\times {10}^{-3}\mathrm{m} \end{gathered}$$

Hence, the value of the beetle’s distance is$6\times {10}^{-3}\mathrm{m}$.