Sound — Question 9

As we know,

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

Total distance travelled by the sound in going and then coming back = 2d

Given,

t₁ = 4 s

t₂ = 6 s

V = 320 ms^-1 and

First echo will be heard from the nearer cliff and the second echo from the farther cliff.

Hence, Distance between cliffs = d₁ + d₂

Therefore, substituting the values in the formula above, we get,

For t₁

$320 = \dfrac{2d_1 }{4} \\[0.5em] \Rightarrow d_1 = \dfrac{320 \times 4}{2} \\[0.5em] \Rightarrow d_1 = 640 m$

So, For t₂

$320 = \dfrac{2d_2}{6} \\[0.5em] \Rightarrow d_2 = \dfrac{320 \times 6}{2} \\[0.5em] \Rightarrow d_2 = 960 m$

Distance between cliffs = d₁ + d₂

= 640 + 960

= 1600 m

Hence, distance between cliffs = 1600 m