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Physics Question on Waves

A transverse wave is described by the equation y=y0sin⁡2π(ft−xλ).y={{y}_{0}}\sin 2\pi \left( ft-\frac{x}{\lambda } \right). The maximum particle velocity is equal to four times the wave velocity, if

A

λ=πy04\lambda =\frac{\pi {{y}_{0}}}{4}

B

λ=πy02\lambda =\frac{\pi {{y}_{0}}}{2}

C

λ=πy0\lambda =\pi {{y}_{0}}

D

λ=2πy0\lambda =2\pi {{y}_{0}}

Answer

λ=πy02\lambda =\frac{\pi {{y}_{0}}}{2}

Explanation

Solution

y=y0sin⁡2π(ft−xλ)y={{y}_{0}}\sin 2\pi \left( ft-\frac{x}{\lambda } \right) .. (i)
For particle velocity, dydt=2πfy0cos⁡2π(ft−xλ)\frac{dy}{dt}=2\pi f{{y}_{0}}\cos 2\pi \left( ft-\frac{x}{\lambda } \right)
Maximum particle velocity, (dydt)max⁡=2πfy0{{\left( \frac{dy}{dt} \right)}_{\max }}=2\pi f{{y}_{0}}
Wave velocity =fλ=f\lambda
Accordingly,
2πfy0=4(fλ)2\pi f{{y}_{0}}=4(f\lambda )
Or λ=πy02\lambda =\frac{\pi {{y}_{0}}}{2}