If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related…

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4 v^2=25-x^2$, then time period is
  1. $\pi$
  2. $2 \pi$
  3. $4 \pi$
  4. $6 \pi$

Solution

$ 4 v^2=25-x^2 $ Differentiating both the side $ \begin{aligned} 4(2 v) \frac{d v}{d t} & =0-2 x \frac{d x}{d t} \\ \frac{d v}{d t} & =A \text { (acceleration) } \\ \frac{d x}{d t} & =v \text { (velocity) } \end{aligned} $ $ \begin{aligned} & \therefore 8 v \times A=-2 x \\ & 4 A=-x \\ & \frac{x}{A}=-\frac{1}{4} \\ & \Rightarrow \quad \frac{\text { Displacement }}{\text { Acceleration }}=\frac{1}{4} \end{aligned} $ Time period $ \begin{aligned} T & =2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}} \\ & =2 \pi \sqrt{\frac{1}{4}}=2 \pi \times \frac{1}{2} \\ & =\pi \mathrm{sec} \end{aligned} $

Asked in: AP EAMCET 2002

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