A) \[2\pi A\]
B) \[\pi A\]
C) \[\frac{\pi A}{2}\]
D) \[\frac{\pi A}{4}\]
Correct Answer: C
Solution :
Key Idea: When particle passes through its equilibrium position, then the velocity is maximum. Velocity of a body changes with displacement \[(y),u=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\]where A is amplitude, \[\omega \] is angular frequency. When particle passes through its equilibrium position, then the velocity is maximum \[{{u}_{\max }}=A\omega \] Given, \[{{u}_{\max }}=4v\] \[\therefore \] \[4v=A\omega \] \[\Rightarrow \] \[\omega =\frac{4v}{A}\] From the relation, speed = frequency \[\times \] wavelength We have \[\text{Wavelength = }\frac{\text{speed}}{\text{frequency}}\] Also, \[\omega \,\,=2\pi n,\,\]where n is frequency Hence, \[n=\frac{\omega }{2\pi }\] Putting this value in Eq. (i), we have \[Wavelength=\frac{v}{\omega /2\pi }=\frac{v}{4v/A2\pi }\] \[\lambda =\frac{\pi A}{2}\]
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