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JEE PHYSICS MCQ: WAVES

1 Medium
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Application +4 | -1
A sine wave has an amplitude AA and a wavelength λ\lambda .Let vv be the wave velocity, and VV be maximum velocity of a particle in the medium
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Correct option: B
V−vV - v,if A=λ/2πA = \lambda /2\pi
Option B (V−vV - v,if A=λ/2πA = \lambda /2\pi) is correct because the wave velocity equals the maximum particle velocity when the amplitude is equal to the wavelength divided by 2π2\pi.

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