Solved Sample Paper 2 — Question 12

(a) 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 = 2 s

t₂ = 5 s

V = 330 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

$330 = \dfrac{2d_1 }{2} \\[1em] \Rightarrow d_1 = \dfrac{330 \times 2}{2} \\[1em] \Rightarrow d_1 = 330 \text{ m}$

For t₂

$330 = \dfrac{2d_2}{5} \\[1em] \Rightarrow d_2 = \dfrac{330 \times 5}{2} \\[1em] \Rightarrow d_2 = 825 \text{ m}$

Distance between cliffs = d₁ + d₂

= 330 + 825

= 1155 m

Hence, distance between cliffs = 1155 m

(b) As we know,

frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$

So, we can say that the natural frequency of vibration of a stretched string is inversely proportional to the radius of the string.

Hence, in order to produce sound waves of different frequencies, strings of different thickness are provided on a stringed instrument.