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Standard XII

Physics

KE & PE Graph

For periodic ...

Question

For periodic motion of small amplitude $A$ , the time period $T$ of this particle is proportional to

A \sqrt{\frac{m}{α}}

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\frac{1}{A} \sqrt{\frac{m}{α}}

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A \sqrt{\frac{α}{m}}

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A \sqrt{\frac{2 α}{m}}

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Solution

The correct option is A $\frac{1}{A} \sqrt{\frac{m}{α}}$
The dimensions of $α$ are $\frac{V (x)}{x^{4}} = \frac{E n e r g y}{{L e n g t h}^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$
It is clear that $\frac{1}{A} \sqrt{\frac{m}{α}}$ has dimensions of time.

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For periodic motion of small amplitude A, the time period T of this particle is proportional to

The correct option is A 1A√mαThe dimensions of α are V(x)x4=EnergyLength4=ML2T−2L4=[ML−2T−2]It is clear that 1A√mα has dimensions of time.

For periodic motion of small amplitude $A$ , the time period $T$ of this particle is proportional to

The correct option is A $\frac{1}{A} \sqrt{\frac{m}{α}}$
The dimensions of $α$ are $\frac{V (x)}{x^{4}} = \frac{E n e r g y}{{L e n g t h}^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$
It is clear that $\frac{1}{A} \sqrt{\frac{m}{α}}$ has dimensions of time.