Q133 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics

Consider a simple harmonic motion (SHM). Let $K$ and $U$ be kinetic energy and potential energy when the displacement in SHM is one-half $\left(\frac{1}{2}\right)$ the amplitude. Which of the correct statement?

  • A
    $\frac{K}{U}=1$
  • B
    $\frac{K}{U}=\frac{1}{2}$
  • C
    $\frac{K}{U}=\frac{4}{3}$
  • Correct answer
    D
    $\frac{K}{U}=3$

Solution

Potential energy of a body performing SHM is given by $U=\frac{1}{2} k x^2$ Here, $x$ is the displacement of the body from mean position. Given, $x=\frac{a}{2}$ [ $a=$ amplitude $]$ $\therefore \quad U=\frac{1}{2} k\left(\frac{a}{2}\right)^2 \Rightarrow U=\frac{1}{2} k \frac{a^2}{4}$ $\ldots(\mathrm{i})$ Now, kinetic energy of this body is given by $K=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(a^2-x^2\right)$ $\Rightarrow \quad K=\frac{1}{2} K\left(a^2-x^2\right)$ Given, $\quad x=\frac{a}{2}$ $\therefore \quad K=\frac{1}{2} K\left(a^2-\frac{a^2}{4}\right) \Rightarrow K=\frac{1}{2} k \times \frac{3 a^2}{4}$ $\ldots(\mathrm{ii})$ From Eqs. (i) and (ii), we get $\frac{K}{U}=\frac{\frac{1}{2} k \times \frac{3 a^2}{4}}{\frac{1}{2} k \times \frac{a^2}{4}}=\frac{3}{1}$

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

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