The speed limit for vehicle on a road is $100 k m / h r$ . A policeman detects a drop of $20$ % in the pitch of the horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit?
(Velocity of sound $= 340 m / s$ ).

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Solution

Given
Allowed speed $= 10 \frac{k m}{h r} = 100 \times \frac{5}{18} = 27.78 \frac{m}{s}$
Speed of listener $V_{L} = 0$
Velocity of sound in air, $v = 340 \frac{m}{s}$
Drop in frequency when approaching and receding $= 20 % = 0.2$
i.e,
$n_{a} (r e c e d i n g) = n_{a} (a p p r o a c h i n g)$ $- 20 % n_{a} (a p p r o a c h i n g)$
$= 0.8 n_{a} (a p p r o a c h i n g)$
$n_{a} = [\frac{V \pm V_{L}}{V \mp V_{S}}] n$
Source is approaching listener. Listener is stationary
$∴ n_{a} (a p p r o a c h i n g) = [\frac{V}{V - V_{S}}] n$ (equation $1$ )
$∴ n_{a} (r e c e d i n g) = [\frac{V}{V + V_{S}}] n$ (equation $2$ )
Dividing equation $2$ by $1$ we get
$∴ \frac{n_{a} (r e c e d i n g)}{n_{a} (a p p r o a c h i n g)}$ $= [\frac{V - V_{S}}{V + V_{S}}]$
$∴ \frac{0.8 \times n_{a} (a p p r o a c h i n g)}{n_{a} (a p p r o a c h i n g)} = [\frac{V - V_{S}}{V + V_{S}}]$
$∴ 0.8 = (\frac{V - V_{S}}{V + V_{S}})$
$∴ 0.8 V + 0.8 V_{S} = V - V_{S}$
$\Rightarrow 0.8 V_{S} + V_{S} = V - 0.8 V$
$\Rightarrow 1.8 V_{S} = 0.2 V$
$\Rightarrow V_{S} = \frac{0.2 V}{1.8} = \frac{0.2 \times 340}{1.8} = 37.77 \frac{m}{s}$
The speed of the car is $37.77 \frac{m}{s}$ which is more than the allowed speed of $27.78 \frac{m}{s}$

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