JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 12)
A transverse wave is represented by $$y=2 \sin (\omega t-k x)\, \mathrm{cm}$$. The value of wavelength (in $$\mathrm{cm}$$) for which the wave velocity becomes equal to the maximum particle velocity, will be :
4$$\pi$$
2$$\pi$$
$$\pi$$
2
Explanation
$${\omega \over k} = A\omega $$
$$ \Rightarrow k = {1 \over A} = {1 \over {2\,cm}}$$
$$ \Rightarrow {{2\pi } \over \lambda } = {1 \over {2\,cm}}$$
$$ \Rightarrow \lambda = 4\pi \,cm$$
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