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A man on the platform is watching two trains, one leaving and the other entering the station with equal speed of \(4 \mathrm{~m} / \mathrm{s}\). If they sound their whistles each of natural frequency \(240 \mathrm{~Hz}\), the number of beats heard by the man (velocity of sound in air \(=320 \mathrm{~m} / \mathrm{s}\) ) will be (A) 6 (B) 3 (C) 0 (D) 12

Short Answer

Expert verified
The number of beats heard by the man is 3. The correct answer is (B) 3.

Step by step solution

01

Determine the Doppler effect formula

The Doppler effect formula for the frequency observed by a stationary observer and a moving source is given by: \(f_o = f_s × \frac{v+v_o}{v-v_s}\) where \(f_o\) is the observed frequency, \(f_s\) is the source frequency (natural frequency of the sound), \(v\) is the velocity of sound in air, \(v_o\) is the velocity of the observer (0 for a stationary observer), \(v_s\) is the velocity of the source (positive if moving towards the observer and negative if moving away).
02

Calculate the apparent frequency of each train

Applying the formula for the Doppler effect, we can find the observed frequencies for both the trains. For the train entering the station towards the man: \(f_{o1} = 240 × \frac{320+0}{320-(-4)}\), \(f_{o1} = 240 × \frac{320}{324}\) For the train leaving the station away from the man: \(f_{o2} = 240 × \frac{320+0}{320-4}\), \(f_{o2} = 240 × \frac{320}{316}\)
03

Calculate the beat frequency

The beat frequency is the absolute difference between the observed frequencies of the two trains: \(f_{beat} = |f_{o1} - f_{o2}|\) Plug in the observed frequencies from Step 2: \(f_{beat} = \left|240 × \frac{320}{324} - 240 × \frac{320}{316}\right|\) Simplify the equation: \(f_{beat} = 240 × \left|\frac{320}{324} - \frac{320}{316}\right|\) Factor out 320 from both terms inside the absolute value: \(f_{beat} = 240 × 320 × \left|\frac{1}{324} - \frac{1}{316}\right|\)
04

Solve for the beat frequency

Now solving for the beat frequency: \(f_{beat} = 240 × 320 × \left|\frac{316-324}{(316)(324)}\right|\) \(f_{beat} = 240 × 320 × \frac{8}{(316)(324)}\) \(f_{beat} = 3\) So, the number of beats heard by the man is 3. The correct answer is (B) 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Beat Frequency
The phenomenon of beat frequency is one that any music lover or musician would find familiar. It occurs when two sound waves of similar, but not identical, frequencies interact with each other. This interaction results in a new sound wave whose frequency is the average of the two original waves and that pulsates or 'beats' at a rate equal to the difference in frequencies of those originating waves. Think of it as the throbbing you hear when two notes that are slightly off-pitch are played simultaneously.

Mathematically, if we have two sound waves with frequencies of \( f_1 \) and \( f_2 \), the beat frequency can be calculated using the formula:
\[ f_{beat} = |f_1 - f_2| \].
When applied to real-life situations such as a man hearing two trains with their whistles at slightly differing frequencies due to Doppler effect, this results in a beat frequency that equals the absolute difference between the higher and the lower observed frequencies of the sound emanating from the trains.
Sound Wave Frequency
Frequency is a cornerstone concept in the study of waves, including sound waves. It is defined as the number of waves (cycles) that pass a given point in one second, and it is measured in hertz (Hz). The frequency of a sound wave directly influences the pitch that we perceive; higher frequencies correspond to higher pitches and vice versa.

The simplest example of a frequency is the natural frequency mentioned in our train scenario, which is the frequency produced by the train's whistle in a stationary position. However, the frequency that the man hears (the observed frequency) may differ from the whistle's natural frequency because of the train's motion altering the wave's frequency as perceived by an observer - an effect known as the Doppler effect.
Doppler Effect Formula
The Doppler effect can make a siren's pitch seem to fall as the ambulance passes us, or it can make a star's light shift toward red or blue, signaling its movement towards or away from us. In physics, it's a wave phenomenon that alters our perception of frequency due to the relative motion between the source and observer.

The formula that represents this effect is fundamental to understanding how the frequency alters:
\[ f_o = f_s \times \frac{v+v_o}{v-v_s} \]
In this formula, \( f_s \) is the natural frequency of the source (like the whistle of the train), \( v \) is the speed of sound in the environment, \( v_o \) is the observer's velocity, and \( v_s \) is the source's velocity. When dealing with a stationary observer and a moving source, the observer's velocity is zero. Applying this formula tells us how the frequency changes and thus allows for the calculation of how we perceive the source's sound.

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Most popular questions from this chapter

A particle of mass \(m\) is moving in a field where the potential energy is given by \(U(x)=U_{0}(1-\cos a x)\), where \(U_{0}\) and a are constants and \(x\) is the displacement from mean position. Then (for small oscillations) (A) The time period is \(T=2 \pi \sqrt{\frac{m}{a U_{0}}}\). (B) The speed of particle is maximum at \(x=0\). (C) The amplitude of oscillations is \(\underline{\pi}\). (D) The time period is \(T=2 \pi \sqrt{\frac{m}{a^{2} U_{0}}}\).

The track followed for two perpendicular SHMs is a perfect ellipse when \(\left(\delta\right.\)-phase difference, \(A_{1}, A_{2}\) amplitudes) (A) \(\delta=\frac{\pi}{4}, A_{1} \neq A_{2}\) (B) \(\delta=\frac{3 \pi}{4}, A_{1} \neq A_{2}\) (C) \(\delta=\frac{\pi}{2}, A_{1} \neq A_{2}\) (D) \(\delta=\pi, A_{1}=A_{2}\)

A whistle producing sound waves of frequencies \(9500 \mathrm{~Hz}\) and above is approaching a stationary person with speed \(v \mathrm{~ms}^{-1}\). The velocity of sound in air is \(300 \mathrm{~ms}^{-1}\). If the person can hear frequencies upto a maximum of \(10,000 \mathrm{~Hz}\), the maximum value of \(\mathrm{v}\) upto which he can hear whistle is (A) \(15 \sqrt{2} \mathrm{~ms}^{-1}\) (B) \(\frac{15}{\sqrt{2}} \mathrm{~ms}^{-1}\) (C) \(15 \mathrm{~ms}^{-1}\) (D) \(30 \mathrm{~ms}^{-1}\)

A bus \(B\) is moving with a velocity \(v_{B}\) in the positive \(x\)-direction along a road as shown in Fig. 9.47. A shooter \(S\) is at a distance \(l\) from the road. He has a detector which can detect signals only of frequency \(1500 \mathrm{~Hz}\). The bus blows horn of frequency \(1000 \mathrm{~Hz}\). When the detector detects a signal, the shooter immediately shoots towards the road along \(S C\) and the bullet hits the bus. Find the velocity of the bullet if velocity of sound in air is \(v=340 \mathrm{~m} / \mathrm{s}\) and \(\frac{v_{B}}{v}=\frac{2}{3 \sqrt{3}}\).

Two particles \(A\) and \(B\) of equal masses are suspended from two massless springs of spring constant \(k_{1}\) and \(k_{2}\), respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of \(A\) and \(B\) is \(\quad\) (A) \(\sqrt{\frac{k_{1}}{k_{2}}}\) (B) \(\frac{k_{2}}{k_{1}}\) (C) \(\sqrt{\frac{k_{2}}{k_{1}}}\) (D) \(\frac{k_{1}}{k_{2}}\)

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