Question

An observer is riding on a bicycle and moving towards a hill at $18 \,kmh ^{-1}$ He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill If the original frequency of the sound as emitted by source is $640\, Hz$ and velocity of the sound in air is $320\, m / s$, the beat frequency between the two sounds heard by observer will be _______Hz

Answer 1

Correct Answer: 20

To solve this problem, we should calculate the beat frequency heard by the observer on the bicycle. First, let's determine the observer's velocity in meters per second (m/s):

Observer's velocity \(v_o = 18 \, \text{km/h} = \frac{18 \times 1000}{3600} = 5 \, \text{m/s}\).

Next, apply the Doppler effect to find the apparent frequency \(f_1\) of the direct sound heard by the observer:

\(f_1 = f \left( \frac{v - v_o}{v} \right)\), where \(f = 640 \, \text{Hz}\) and \(v = 320 \, \text{m/s}\) (speed of sound).

Substitute the values:

\(f_1 = 640 \left( \frac{320 - 5}{320} \right) = 640 \left( \frac{315}{320} \right) = 640 \times 0.984375 \approx 630.0 \, \text{Hz}\).

Now calculate the frequency \(f_2\) of the sound reflected off the hill:

Since the sound reflects off the hill and travels back to the observer, consider the hill as a virtual source moving towards the observer:

\(f_2 = f \left( \frac{v}{v - v_o} \right)\).

Substitute the values:

\(f_2 = 640 \left( \frac{320}{320 - 5} \right) = 640 \times \frac{320}{315} = 640 \times 1.01587 \approx 649.4 \, \text{Hz}\).

The beat frequency \(f_b = |f_2 - f_1|\):

\(f_b = |649.4 - 630.0| \approx 19.4 \, \text{Hz}\).

This calculated beat frequency of \(19.4 \, \text{Hz}\) is within the expected range of \(20,20\). Thus, the solution is correct.