TS EAMCET · Physics · Waves and Sound
Consider two tuning forks with natural frequency \(250 \mathrm{~Hz}\). One is moving away and another is moving towards a stationary observer at same speed. If the observer hears beats of frequency \(5 \mathrm{~Hz}\), then the speed of the tuming fork is : (Given, speed of sound wave is \(350 \mathrm{~m} / \mathrm{s}\).)
- A \(2.5 \mathrm{~m} / \mathrm{s}\)
- B \(3.5 \mathrm{~m} / \mathrm{s}\)
- C \(5.0 \mathrm{~m} / \mathrm{s}\)
- D \(2.0 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(3.5 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Given, speed of sound, \(v=350 \mathrm{~m} / \mathrm{s}\), actual frequency, \(n_0=250 \mathrm{~Hz}\) and number of beats heared, \(x=5\) As source is moving towards the observer therefore, the apparent frequency, \[ n_1=n_0 \cdot \frac{v}{v-v_s} \] Similarly, as source is…
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