Q122 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
A particle of mass '$m$' is executing S.H.M. about the origin on x-axis with frequency $\sqrt{\dfrac{Ka}{\pi m}}$, where K is a constant and a is the amplitude of S.H.M. If '$x$' is the displacement of a particle at time '$t$', the potential energy of a particle will be
- A$\dfrac{1}{2}Kax^2$
- B$\pi Kax^2$
- Correct answerC$2\pi Kax^2$
- D$2Kax^2$
Solution
The frequency of the particle executing S.H.M. is given by $f = \sqrt{\dfrac{Ka}{\pi m}}$
The angular frequency is $\omega = 2\pi f = 2\pi \sqrt{\dfrac{Ka}{\pi m}}$
Squaring both sides, we get $\omega^2 = 4\pi^2 \left(\dfrac{Ka}{\pi m}\right) = \dfrac{4\pi Ka}{m}$
The potential energy of a particle in S.H.M. at a displacement $x$ is given by $U = \dfrac{1}{2} m \omega^2 x^2$
Substituting the value of $\omega^2$, we get $U = \dfrac{1}{2} m \left(\dfrac{4\pi Ka}{m}\right) x^2$
$U = 2\pi K a x^2$
Answer: $2\pi Kax^2$
The angular frequency is $\omega = 2\pi f = 2\pi \sqrt{\dfrac{Ka}{\pi m}}$
Squaring both sides, we get $\omega^2 = 4\pi^2 \left(\dfrac{Ka}{\pi m}\right) = \dfrac{4\pi Ka}{m}$
The potential energy of a particle in S.H.M. at a displacement $x$ is given by $U = \dfrac{1}{2} m \omega^2 x^2$
Substituting the value of $\omega^2$, we get $U = \dfrac{1}{2} m \left(\dfrac{4\pi Ka}{m}\right) x^2$
$U = 2\pi K a x^2$
Answer: $2\pi Kax^2$
Hence, the correct answer is option: 'C'.