When a vibrating tuning fork moves towards a stationary observer with a speed of 50 m/s, the observer hears a frequency of 350 Hz. The frequency of vibration of the fork is:(Take speed of sound = 350 m/s)
When a vibrating tuning fork moves towards a stationary observer with a speed of 50 m/s, the observer hears a frequency of 350 Hz. The frequency of vibration of the fork is:(Take speed of sound = 350 m/s)
350 Hz
400 Hz
200 Hz
300 Hz
250 Hz
The Correct Option is D
Approach Solution - 1
Given:
- Speed of tuning fork (source), \( v_s = 50 \, \text{m/s} \) (moving towards observer)
- Observed frequency, \( f' = 350 \, \text{Hz} \)
- Speed of sound, \( v = 350 \, \text{m/s} \)
- Observer is stationary (\( v_o = 0 \))
Step 1: Apply Doppler Effect Formula
When a sound source moves towards a stationary observer, the observed frequency is given by:
\[ f' = \left( \frac{v}{v - v_s} \right) f \]
where \( f \) is the actual frequency of the tuning fork.
Step 2: Solve for Actual Frequency (\( f \))
Rearrange the formula to solve for \( f \):
\[ f = f' \left( \frac{v - v_s}{v} \right) \]
Substitute the known values:
\[ f = 350 \left( \frac{350 - 50}{350} \right) \]
\[ f = 350 \left( \frac{300}{350} \right) \]
\[ f = 350 \times \frac{300}{350} = 300 \, \text{Hz} \]
Conclusion:
The actual frequency of vibration of the tuning fork is 300 Hz.
Answer: \(\boxed{D}\)
Approach Solution -2
Step 1: Recall the Doppler effect formula for sound.
The observed frequency \( f' \) when a source moves towards a stationary observer is given by:
\[ f' = f \frac{v}{v - v_s}, \]
where:
- \( f' \) is the observed frequency,
- \( f \) is the actual frequency of the source,
- \( v \) is the speed of sound in air, and
- \( v_s \) is the speed of the source relative to the medium (air).
We are given:
- \( f' = 350 \, \text{Hz}, \)
- \( v = 350 \, \text{m/s}, \)
- \( v_s = 50 \, \text{m/s}. \)
Step 2: Solve for the actual frequency \( f \).
Rearranging the formula for \( f \):
\[ f = f' \cdot \frac{v - v_s}{v}. \]
Substitute the given values:
\[ f = 350 \cdot \frac{350 - 50}{350}. \]
Simplify:
\[ f = 350 \cdot \frac{300}{350} = 350 \cdot \frac{6}{7} = 300 \, \text{Hz}. \]
Final Answer: The frequency of vibration of the fork is \( \mathbf{300 \, \text{Hz}} \), which corresponds to option \( \mathbf{(D)} \).
Top Questions on Waves
- In an open organ pipe \( \nu_3 \) and \( \nu_6 \) are 3rd and 6th harmonic frequencies, respectively. If \( \nu_6 - \nu_3 = 2200\ \text{Hz} \), then the length of the pipe is _________ mm. (Take velocity of sound in air as \(330\ \text{m s}^{-1}\).)
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(Speed of sound in air is \(324 \, \text{m/s}\) and \( \sqrt{5} = 2.23 \))
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Concepts Used:
Beats
What is Beat Frequency?
Let’s see from fig.m, the frequency of a pink-colored wave is f1, and that of a green-colored wave is f2. So, the frequency of the beat is the difference between these two, which is:
fBEATS = |f1 - f2|
Interference and Beats
When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.
What is Interference?
When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.
The Application of Beats:
- Beats are used in determining the unknown frequency
- Beats are used in determining the existence of poisonous gases in mines.