MCQ
A transverse wave is represented by $y=2 \sin$ $(\omega t - kx ) cm$. The value of wavelength (in $cm$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be.
- ✓$4 \pi$
- B$2 \pi$
- C$\pi$
- D$2$
✓
Answer
Correct option: A.
$4 \pi$
a
$y=2 \sin (\omega t-k x)$
$y=2 \sin (\omega t-k x)$
Maximum particle velocity $= A \omega$
Wave velocity $=\frac{\omega}{ k }$
$\frac{\omega}{ k }= A \omega$
$k =\frac{1}{ A }=\frac{2 \pi}{\lambda}$
$\lambda =2 \pi A$
$=4 \pi\,cm$
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