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JEE Mains · Physics · STD 11 - 14. waves and sound
In a transverse wave the distance between a crest and neighbouring trough at the same instant is \(4.0\, cm\) and the distance between a crest and trough at the same place is \(1 .0\, cm\) . The next crest appears at the same place after a time interval of \(0.4\,s\). The maximum speed of the vibrating particles in the medium is
- A \(\frac{{3\pi }}{2}\,cm/s\)
- B \(\frac{{5\pi }}{2}\,cm/s\)
- C \(\frac{{\pi }}{2}\,cm/s\)
- D \(2\pi \,cm/s\)
Answer & Solution
Correct Answer
(B) \(\frac{{5\pi }}{2}\,cm/s\)
Step-by-step Solution
Detailed explanation
Transverse wave equation \(y=a \sin (\omega t+k x)\) Next crest appears at time interval of \(0.4 \mathrm{s}\) So, the time period \(T=0.4\) \(\therefore\) Angular Frequency \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.4}\) \(\therefore\) Speed of the particle…
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