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  • In a transverse wave the distance between a crest and neighbouring trough at the same instant is  and the distance

In a transverse wave the distance between a crest and neighbouring trough at the same instant is 4.0cm and the distance between a crest and trough at the same place is 1.0cm. The next crest appears at the same place after a time interval of 0.4s. The maximum speed of the vibrating particles in the medium is : 

Option: 1

\frac{3\pi }{2}\: cm/s
 


Option: 2

\frac{5\pi }{2}\: cm/s


Option: 3

\frac{\pi }{2}\: cm/s

 


Option: 4

2\pi \: cm/s


Answers (1)

best_answer

 

 

 

 

Transverse wave equation

y=a \sin (\omega t+k x)

Next crest appears at time interval of 0.4s
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

 \begin{array}{l} using \ \ a=1 / 2 c m \\ v_{max}=\frac{1}{2} \times \frac{2 \pi}{0.4}=\frac{2 \pi}{8} \times 10 \\ v_{max}=\frac{5 \pi}{2} \mathrm{cm} / \mathrm{s} \end{array}

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Nehul

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