A stationary source is emitting sound at a fixed frequency $\mathrm{f}_{0}$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of $\mathrm{f}_{0}$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is $330 \mathrm{~ms}^{-1}$.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Waves & Sound

A stationary source is emitting sound at a fixed frequency

f_{0}

, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of

f_{0}

. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is

330 ms^{- 1}

Answer: 7

Solution

When a car moving with speed

v

approaches a stationary source, the frequency it receives is:

f_{received} = f_{0} \frac{c + v}{c}

This frequency is then reflected back, and the source (stationary) receives:

f_{reflected} = f_{received} \frac{c}{c - v} = f_{0} \frac{c + v}{c - v}

For small

v ≪ c

f_{reflected} \approx f_{0} (1 + \frac{2 v}{c})

The difference in reflected frequencies from two cars with speeds

v_{1}

and

v_{2}

is:

Δ f = f_{0} \frac{2 ( v _{1} - v _{2} )}{c}

Given

Δ f = 0.012 f_{0}

0.012 f_{0} = f_{0} \frac{2Δ v}{c}

Δ v = \frac{0.012 \times 330}{2} = 1.98 m/s

Converting to km/hr:

Δ v = 1.98 \times 3.6 = 7.128 \approx 7 km/hr

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Waves & Sound

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