Q131 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Physics

A particle undergoing simple harmonic motion has an amplitude of $10 \mathrm{~cm}$. When the particle is at a displacement of $6 \mathrm{~cm}$ from the centre, then the ratio of its kinetic energy to potential energy is

  • A
    $3: 2$
  • B
    $9: 4$
  • Correct answer
    C
    $16: 9$
  • D
    $4: 3$

Solution

Amplitude of particle performing SHM, $$ A=10 \mathrm{~m} $$ Instantaneous displacement of the particle, $$ x=6 \mathrm{~cm} $$ Kinetic energy of the particle, $$ K=\frac{1}{2} m \omega^2\left(A^2-x^2\right) $$ Potential energy of particle, $$ \begin{aligned} U & =\frac{1}{2} m \omega^2 x^2 \\ \frac{K}{U} & =\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \\ & =\frac{A^2-x^2}{x^2}=\frac{A^2}{x^2}-1 \\ & =\left(\frac{A}{x}\right)^2-1=\left(\frac{10}{6}\right)^2-1=\frac{25}{9}-1 \\ \Rightarrow \quad \frac{K}{U} & =\frac{16}{9} \Rightarrow K: U=16: 9 \end{aligned} $$

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

Relevant Questions