JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2016
- Easy
- Numerical
Two engines pass each other moving in opposite directions with uniform speed of \(30\,m/s\) . One of them is blowing a whistle of frequency \(540\,Hz.\) Calculate the frequency heard by driver of second engine before they pass each other ... \(Hz\). Speed of sound is \(330\,m/sec\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(648\)
Step-by-step Solution
Detailed explanation
We know that the apperent frequency
\(\mathrm{f}^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{0}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}\) from Doppler's effect
where \(v_{0}=v_{s}=30 \mathrm{m} / \mathrm{s},\) velocity of observer and source
Speed of sound \(v=330 \mathrm{m} / \mathrm{s}\)
\(\because\) Frequency of whistle \((\mathrm{f})=540 \mathrm{Hz}\)
\(\therefore \mathrm{f}^{\prime}=\frac{330+30}{330-30} \times 540=648 \mathrm{Hz}\)
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