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Question

Physics Question on Oscillations

A sine wave has an amplitude A and wavelength λ\lambda . Let V be the wave velocity and maximum velocity of a particle in the medium. Then

A

V=υI2πλV=\upsilon\,I\, 2\pi\lambda

B

V=υifλ=3A2πV=\upsilon\,if\, \lambda=\frac{3A}{2\pi}

C

V cannot be equal to υ\upsilon

D

V=υifA=λ2πV=\upsilon\,if\, A=\frac{\lambda}{2\pi}

Answer

V=υifA=λ2πV=\upsilon\,if\, A=\frac{\lambda}{2\pi}

Explanation

Solution

Using υp=2πAλυω\upsilon_p = \frac{2\pi A}{\lambda} \upsilon_{\omega} when υp=υω\upsilon_p = \upsilon_{\omega},
we get, A = λ2π\frac{\lambda}{2 \pi}.