Question

A man is watching two trains, one leaving and the other coming in with equal speeds of 4$m{s^{ - 1}}$ . If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man is:(velocity of sound in air = 320$m{s^{ - 1}}$)
A) 6
B) 3
C) 0
D) 12

Answer 1

In order to solve the question one need to know about the effect of Doppler and its formula which is $f' = \dfrac{{\left( {v + {v_0}} \right)}}{v}f$ where, ${f'}$ is the observed frequency, f is the actual frequency, v is the velocity of sound waves in air = 320$m{s^{ - 1}}$and ${v_0}$is the velocity of the observer. We will calculate for both the trains and subtract them for beats.

Step by step solution:
Step 1:
Before solving the question let us see some definition and formulas that are important to solve the problems
Doppler’s Effect: Doppler's Effect. Doppler Effect is the phenomenon of motion-related frequency change. Consider if a truck is coming from very far off location as it approaches near our house, the sound increases and when it passes our house the sound will be maximum.
The formula of the Doppler’s Effect is $f' = \dfrac{{\left( {v + {v_0}} \right)}}{v}f$ where, f’ is the observed frequency, f is the actual frequency, v is the velocity of sound waves in air = 320$m{s^{ - 1}}$and ${v_0}$is the velocity of the observer.
Step 2:
Now coming to the question:
We are given:
Observer is stationary so ${v_0} = 0$
A man is watching two trains, one leaving and the other coming in with equal speeds of 4$m{s^{ - 1}}$=${u_S}$ , here ${u_S}$ is the velocity of the source.
Sound of their whistles, each of frequency 240 Hz=${h_0}$
Sound waves in air = 320$m{s^{ - 1}}$=v
Let the number of beats be $\Delta n$=${n_1} - {n_2}$ where, ${n_1}$is the apparent frequency heard by man from approaching train, and ${n_2}$ is the apparent frequency heard by man from receding train.
Now using the Doppler’s effect formula $n' = n\left( {\dfrac{{v \pm {v_0}}}{{v \pm {u_S}}}} \right)$
Calculating for ${n_1}$=${n_0}\left( {\dfrac{{v \pm {v_0}}}{{v - {u_S}}}} \right)$ , there is minus sign in denominator because the approaching train will have more frequency than the receding train.
Solving for ${n_1}$ substituting the value, $240\left( {\dfrac{{320}}{{320 - 4}}} \right)$ …… (1)
Now calculating same for ${n_2}$=${n_0}\left( {\dfrac{{v \pm {v_0}}}{{v + {u_S}}}} \right)$ , this time there is positive sign in denominator because source is going far from observer so apparent frequency will be less than actual frequency.
Then ${n_2}$ equal to $240\left( {\dfrac{{320}}{{320 + 4}}} \right)$ ……. (2)
Putting 1 and 2 in the formula of $\Delta n$=$240 \times 320\left( {\dfrac{1}{{316}} - \dfrac{1}{{324}}} \right)$ which on solving give $\Delta n$ =6 Hz
Hence, the number of beats heard by the man is 6 Hz

Option A is correct.

Note: In this question if a person know about the formula of Doppler’s effect than this is an easy question. But one needs to be caution before solving the numerical part as there is a high chance of mistakes while solving the mathematical terms. One mistake in between the equation will mislead from getting the solutions.