Sound — Question 8

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

5 vibrations are produced in 1 second.

Therefore, time for 8 vibrations = ($\dfrac{1}{5}$ x 8) s= $\dfrac{8}{5}$ s = 1.6 s

Hence, t = 1.6 s

V = 340 ms^-1

Substituting the values in the formula above, we get,

$340 = \dfrac{2d}{1.6} \\[0.5em] \Rightarrow d = \dfrac{340 \times 1.6}{2} \\[0.5em] \Rightarrow d = 340 \times 0.8 \\[0.5em] \Rightarrow d = 272 m$

Therefore, the distance between the cliff and the observer = 272 m