MHT CET · Physics · Waves and Sound
The velocity of sound is \(340 \mathrm{~m} / \mathrm{s}\). A source of sound having frequency of \(90 \mathrm{~Hz}\) is moving towards a stationary observer with a speed of one-tenth that of sound. The apparent frequency of sound as heard by the observer is
- A \(45 \mathrm{~Hz}\)
- B \(100 \mathrm{~Hz}\)
- C \(80 \mathrm{~Hz}\)
- D \(50 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(100 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Apparent frequency of sound heard by the observer is:
\(f=f_0\left(\frac{v}{v-v_s}\right)\)
Where, \(v=340 \mathrm{~m} / \mathrm{s}\) is the speed of sound \& \(v_s=v / 10\) is the speed of the source morning towards the observer.
\(\therefore f=90\left(\frac{v}{v-\frac{v}{10}}\right) \mathrm{Hz}=90\left(\frac{10}{9}\right) \mathrm{Hz}=100 \mathrm{~Hz}\)
\(f=f_0\left(\frac{v}{v-v_s}\right)\)
Where, \(v=340 \mathrm{~m} / \mathrm{s}\) is the speed of sound \& \(v_s=v / 10\) is the speed of the source morning towards the observer.
\(\therefore f=90\left(\frac{v}{v-\frac{v}{10}}\right) \mathrm{Hz}=90\left(\frac{10}{9}\right) \mathrm{Hz}=100 \mathrm{~Hz}\)
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