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JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2013
- Easy
- MCQ
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
Pick an option to check your answer.
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
\(\nu=\frac{d y}{d t}=a \omega \sin (\omega t+k x)\)
\(\nu_{\max }=a \omega\)
\(a=1 / 2 c m\)
\(=\frac{1}{2} \times \frac{2 \pi}{0.4}=\frac{2 \pi}{8} \times 10\)
\(=\frac{5 \pi}{2} \mathrm{cm} / \mathrm{s}\)
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