TS EAMCET · Physics · Waves and Sound
A source and an observer move away from each other with same velocity of \(10 \mathrm{~ms}^{-1}\) with respect to ground. If the observer finds the frequency of sound coming from the source as \(1980 \mathrm{~Hz}\), then actual frequency of the source is (speed of sound in air \(=340 \mathrm{~ms}^{-1}\) )
- A \(1950 \mathrm{~Hz}\)
- B \(2100 \mathrm{~Hz}\)
- C \(2132 \mathrm{~Hz}\)
- D \(2486 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(2100 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
By Doppler effect formula \[ f=f_o\left(\frac{v-v_0}{v+v_s}\right) \] Here \(\mathrm{f}=1980 \mathrm{~Hz}\) Speed of sound in air, \(v=340 \mathrm{~m} / \mathrm{s}\)…
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