Q155 / 200 · TS EAMCET 2002 · Physics
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
- Correct answerA$\pi$
- B$2 \pi$
- C$4 \pi$
- D$6 \pi$
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
$$
\begin{aligned}
T & =2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}} \\
& =2 \pi \sqrt{\frac{1}{4}}=2 \pi \times \frac{1}{2} \\
& =\pi \mathrm{sec}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.