Q178 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
If the kinetic energy of simple pendulum is $\dfrac{1}{4}$ of its total energy, then the displacement ($x$) and amplitude ($A$) are related as
- A$x = A$
- B$x = 2A$
- Correct answerC$x = \dfrac{\sqrt{3}}{2}A$
- D$x = \dfrac{\sqrt{3}}{4}A$
Solution
Total energy of a simple pendulum executing SHM is given by $E = \dfrac{1}{2} m \omega^2 A^2$
Kinetic energy at displacement $x$ is given by $K = \dfrac{1}{2} m \omega^2 (A^2 - x^2)$
Given that $K = \dfrac{1}{4} E$
$\dfrac{1}{2} m \omega^2 (A^2 - x^2) = \dfrac{1}{4} \left( \dfrac{1}{2} m \omega^2 A^2 \right)$
$A^2 - x^2 = \dfrac{A^2}{4}$
$x^2 = A^2 - \dfrac{A^2}{4} = \dfrac{3A^2}{4}$
$x = \dfrac{\sqrt{3}}{2} A$
Answer: $x = \dfrac{\sqrt{3}}{2}A$
Kinetic energy at displacement $x$ is given by $K = \dfrac{1}{2} m \omega^2 (A^2 - x^2)$
Given that $K = \dfrac{1}{4} E$
$\dfrac{1}{2} m \omega^2 (A^2 - x^2) = \dfrac{1}{4} \left( \dfrac{1}{2} m \omega^2 A^2 \right)$
$A^2 - x^2 = \dfrac{A^2}{4}$
$x^2 = A^2 - \dfrac{A^2}{4} = \dfrac{3A^2}{4}$
$x = \dfrac{\sqrt{3}}{2} A$
Answer: $x = \dfrac{\sqrt{3}}{2}A$
Hence, the correct answer is option: 'C'.