Q131 / 160 · AP EAMCET 2015 · Physics

The amplitude of a simple pendulum is $10 \mathrm{~cm}$. When the pendulum is at a displacement of $4 \mathrm{~cm}$ from the mean position, the ratio of kinetic and potential energies at that point is

  • Correct answer
    A
    $5.25$
  • B
    $2.5$
  • C
    $4.5$
  • D
    $7.5$

Solution

$\begin{aligned} & \text { } \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\ & \mathrm{P} . \mathrm{E}=\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2 \end{aligned}$ Therefore, the ratio $\begin{aligned} & \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \Rightarrow \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\mathrm{A}^2-\mathrm{x}^2}{\mathrm{x}^2} \\ & \Rightarrow \frac{10^2-4^2}{4^2}=\frac{84}{16} \\ & \Rightarrow \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{21}{4}=5.25 \end{aligned}$

Hence, the correct answer is option: 'A'.

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