A source, approaching with speed $u$ towards the open end of a stationary pipe of length $L$, is emitting a sound of frequency $f _{ s }$. The farther end of the pipe is closed .The speed of sound in air is $v$ and $f _{0}$ is the fundamental frequency of the pipe. For which of the following combination(s) of $u$ and $f_{s}$, will the sound reaching the pipe lead to a resonance?
- $u =0.8 \,v$ and $f _{ s }= f _{0}$
- $u=0.8\, v$ and $f _{ s }=2\, f _{0}$
- $u =0.8 \,v$ and $f _{ s }=0.5\, f _{0}$
- $u=0.5 \,v$ and $f _{ s }=1.5 \,f _{0}$
The Correct Option is D
Solution and Explanation
The given problem involves a source approaching a stationary pipe with an open end and a closed end. The source emits a sound at frequency \( f_s \), and the speed of sound in air is \( v \). The length of the pipe is \( L \), and the fundamental frequency of the pipe is \( f_0 \). We are asked to determine the combination of the speed of the source \( u \) and the emitted frequency \( f_s \) that will lead to resonance in the pipe.
Step 1: Doppler Effect
Since the source is moving towards the pipe, we need to apply the Doppler effect to calculate the frequency of the sound as it reaches the open end of the pipe. The Doppler-shifted frequency \( f' \) at the open end of the pipe, when the source is moving towards the observer, is given by:
\( f' = \frac{f_s}{1 - \frac{u}{v}} \)
Where \( f_s \) is the frequency of the source, \( u \) is the speed of the source, and \( v \) is the speed of sound in air.
Step 2: Resonance Condition
The fundamental frequency of the pipe, \( f_0 \), is determined by the length \( L \) of the pipe, where the closed end causes the air to reflect and set up a standing wave. For a pipe with one closed end and one open end, the fundamental frequency \( f_0 \) is given by:
\( f_0 = \frac{v}{4L} \)
For resonance to occur, the frequency of the sound reaching the pipe should be a harmonic of the fundamental frequency \( f_0 \). The first resonance occurs when the frequency \( f' \) is equal to the fundamental frequency \( f_0 \). Therefore, for the first resonance to happen, we set:
\( f' = f_0 \)
Step 3: Solving for \( u \) and \( f_s \)
Substituting the equation for \( f' \) and setting it equal to \( f_0 \), we get:
\( \frac{f_s}{1 - \frac{u}{v}} = \frac{v}{4L} \)
Rearranging this equation, we find:
\( f_s = \frac{v}{4L} \left(1 - \frac{u}{v} \right) \)
Step 4: Considering the harmonic
The first resonance condition occurs when \( f_s \) is equal to \( 1.5 f_0 \) (since the first resonance in a pipe with one closed end is at 3 times the fundamental frequency). Thus, we have:
\( f_s = 1.5 f_0 = 1.5 \times \frac{v}{4L} \)
Substituting this into the equation for \( f_s \) from Step 3, we get:
\( 1.5 \times \frac{v}{4L} = \frac{v}{4L} \left( 1 - \frac{u}{v} \right) \)
Canceling the common terms, we get:
\( 1.5 = 1 - \frac{u}{v} \)
Solving for \( u \), we get:
\( \frac{u}{v} = 0.5 \)
Thus, \( u = 0.5v \).
Final Answer:
The combination of \( u = 0.5v \) and \( f_s = 1.5 f_0 \) will lead to resonance.
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Concepts Used:
Waves
Waves are a disturbance through which the energy travels from one point to another. Most acquainted are surface waves that tour on the water, but sound, mild, and the movement of subatomic particles all exhibit wavelike properties. inside the most effective waves, the disturbance oscillates periodically (see periodic movement) with a set frequency and wavelength.
Types of Waves:
Transverse Waves -
Waves in which the medium moves at right angles to the direction of the wave.
Examples of transverse waves:
- Water waves (ripples of gravity waves, not sound through water)
- Light waves
- S-wave earthquake waves
- Stringed instruments
- Torsion wave
The high point of a transverse wave is a crest. The low part is a trough.
Longitudinal Wave -
A longitudinal wave has the movement of the particles in the medium in the same dimension as the direction of movement of the wave.
Examples of longitudinal waves:
- Sound waves
- P-type earthquake waves
- Compression wave