A person standing between two vertical cliffs produces a sound and hears the first echo after 3 s. If the distance between the cliffs is 1760 m, then after how much time will he hear the second echo? (The speed of sound = 320 ms-1)
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Given,
- Time after which the first echo is heard (t1) = 3 s
- Distance between the two cliffs = 1760 m
- Speed of sound (v) = 320 m s-1
The first echo is heard from the nearer cliff.
Let the distance of the person from the nearer cliff be d1.
$$\begin{aligned}\text{Speed of sound} = \frac{2\text d_1}{\text t_1} \\[1em] \Rightarrow 2\text d_1 = \text v \times \text t_1 = 320 \times 3 = 960\ \text m \\[1em] \Rightarrow \text d_1 = \frac{960}{2} = 480\ \text m\end{aligned}$$
Then,
Distance of the person from the farther cliff (d2) = 1760 - 480 = 1280 m
Let the second echo be heard after time t2.
$$\begin{aligned}\text t_2 = \frac{2\text d_2}{\text v} = \frac{2 \times 1280}{320} \\[1em] = \frac{2560}{320} = 8\ \text s\end{aligned}$$
Hence, he will hear the second echo 8 s after producing the sound, i.e., 5 s after the first echo.