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A transverse wave is represented by \mathrm{y=2 \sin (\omega t-k x) \mathrm{cm}. }The value of wavelength \mathrm{(in \; \mathrm{cm}) } for which the wave velocity becomes equal to the maximum particle velocity, will be :

Option: 1

4 \pi


Option: 2

2 \pi


Option: 3

\pi


Option: 4

2


Answers (1)

best_answer

\mathrm{y=2 \sin (\omega t-k x)}
Wave velocity = Maximum particle velocity

\mathrm{\frac{\omega}{k}=A \omega }

\mathrm{\frac{1}{(2 \pi / \lambda)}=A }

\mathrm{\lambda=A(2 \pi) }

\mathrm{\lambda=4 \pi}

Hence (1) is correct option.





 

Posted by

Pankaj Sanodiya

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