AP EAMCET · PHYSICS · Waves and Sound
An observer and a source emitting sound of frequency \(120 \mathrm{~Hz}\) are on the \(X\)-axis. The observer is stationary while the source of sound is in motion given by the equation \(x=3 \sin \omega t\) ( \(x\) is in metres and \(t\) is in seconds). If the difference between the maximum and minimum frequencies of the sound observed by the observers is \(22 \mathrm{~Hz}\), then the value of \(\omega\) is ( ppeed of sound in air \(=330 \mathrm{~ms}^{-1}\) )
- A \(33 \mathrm{rad} \mathrm{s}^{-1}\)
- B \(36 \mathrm{rad} \mathrm{s}^{-1}\)
- C \(20 \mathrm{rad} \mathrm{s}^{-1}\)
- D \(10 \mathrm{rad} \mathrm{s}^{-1}\)
Answer & Solution
Correct Answer
(D) \(10 \mathrm{rad} \mathrm{s}^{-1}\)
Step-by-step Solution
Detailed explanation
Instantaneous speed of source is \(v=\frac{d x}{d t}=3 \omega \sin \omega t\) Difference between maximum and minimum frequencies is \(22 \mathrm{~Hz}\). So, \(f_{\max }-f_{\min }=f\left(\frac{v}{v-v_s}\right)-f\left(\frac{v}{v+v_s}\right)=22\)…
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