A) \[\frac{{{\beta }^{2}}}{{{\alpha }^{2}}}\]
B) \[\frac{\alpha }{\beta }\]
C) \[\frac{{{\beta }^{2}}}{\alpha }\]
D) \[\frac{2\pi \beta }{\alpha }\]
Correct Answer: D
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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