A) \[\frac{{{\beta }^{2}}}{{{\alpha }^{2}}}\]
B) \[\frac{\alpha }{\beta }\]
C) \[\frac{{{\beta }^{2}}}{\alpha }\]
D) \[\frac{2\pi \beta }{\alpha }\]
Correct Answer: B
Solution :
For a particle executing SHM, we have maximum acceleration, \[\alpha =A{{\omega }^{2}}\] ?(i) where, A is maximum amplitude and co is angular velocity of a particle. Maximum velocity, \[\beta =A\omega \,\] ?(ii) Comparing Eq. (i) and Eq. (ii), we get \[\frac{\alpha }{\beta }=\frac{A{{\omega }^{2}}}{A\omega }\Rightarrow \frac{\alpha }{\beta }=\omega =\frac{2\pi }{T}\] i.e. \[T=\frac{2\pi \beta }{\alpha }\] Thus, its time period of vibration, \[T=\frac{2\pi \beta }{\alpha }\]
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