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Equation of SHM is given by y(x,t) = a_{o} \sin2\pi(vt -\frac{x}{\lambda }). If the maximum particle velocity is three times the wave velocity, the wavelength \lambda of the wave is 

  • Option 1)

    \frac{\pi a_{o}}{3}

  • Option 2)

    \frac{2\pi a_{o}}{3}

  • Option 3)

    \pi a_{o}

  • Option 4)

    \frac{\pi a_{o}}{2}

 

Answers (1)

best_answer

v_{p} = a_{o} \omega = 2\pi a_{o}v

wave velocity = v \lambda

 given , 2\pi a_{o}v = 3 v \lambda

\lambda = \frac{2\pi a_{o} } {3}

 

Relation between phase velocity and wave speed -

V_{P}= -V\: \frac{dy}{dx}

- wherein

V_{P}= particle velocity

V = wave velocity

\frac{dy}{dx}= slope of curve

 

 

 


Option 1)

\frac{\pi a_{o}}{3}

This is incorrect.

Option 2)

\frac{2\pi a_{o}}{3}

This is correct.

Option 3)

\pi a_{o}

This is incorrect.

Option 4)

\frac{\pi a_{o}}{2}

This is incorrect.

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