Q112 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
The amplitude of a particle executing SHM is $3$ cm. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is (in cm)
- A$1$
- Correct answerB$2$
- C$3$
- D$4$
Solution
Given amplitude $A = 3$ cm.
The kinetic energy of a particle in SHM is given by $K = \dfrac{1}{2} m \omega^2 (A^2 - x^2)$ and the potential energy is $U = \dfrac{1}{2} m \omega^2 x^2$.
Given that kinetic energy is $25\%$ more than potential energy:
$K = U + \dfrac{25}{100} U = \dfrac{5}{4} U$
$\dfrac{1}{2} m \omega^2 (A^2 - x^2) = \dfrac{5}{4} \left( \dfrac{1}{2} m \omega^2 x^2 \right)$
$A^2 - x^2 = \dfrac{5}{4} x^2$
$A^2 = \dfrac{9}{4} x^2$
$x = \dfrac{2}{3} A$
Substituting $A = 3$ cm:
$x = \dfrac{2}{3} \times 3 = 2$ cm.
Answer: $2$
The kinetic energy of a particle in SHM is given by $K = \dfrac{1}{2} m \omega^2 (A^2 - x^2)$ and the potential energy is $U = \dfrac{1}{2} m \omega^2 x^2$.
Given that kinetic energy is $25\%$ more than potential energy:
$K = U + \dfrac{25}{100} U = \dfrac{5}{4} U$
$\dfrac{1}{2} m \omega^2 (A^2 - x^2) = \dfrac{5}{4} \left( \dfrac{1}{2} m \omega^2 x^2 \right)$
$A^2 - x^2 = \dfrac{5}{4} x^2$
$A^2 = \dfrac{9}{4} x^2$
$x = \dfrac{2}{3} A$
Substituting $A = 3$ cm:
$x = \dfrac{2}{3} \times 3 = 2$ cm.
Answer: $2$
Hence, the correct answer is option: 'B'.