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NEET PHYSICSOscillationsMedium

Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is:

A

\frac{\sqrt{5}}{\pi}

B

\frac{\sqrt{5}}{2\pi}

C

\frac{4\pi}{\sqrt{5}}

D

\frac{2\pi}{\sqrt{3}}

Step-by-Step Solution

In Simple Harmonic Motion (SHM), the magnitude of velocity (vv) and acceleration (aa) at a displacement xx from the mean position are given by: v=ωA2−x2v = \omega \sqrt{A^2 - x^2} a=ω2xa = \omega^2 x where AA is the amplitude and ω\omega is the angular frequency.

Given that the magnitude of velocity equals the magnitude of acceleration (∣v∣=∣a∣|v| = |a|) at x=2 cmx = 2 \text{ cm} and A=3 cmA = 3 \text{ cm}: ωA2−x2=ω2x\omega \sqrt{A^2 - x^2} = \omega^2 x

Dividing both sides by ω\omega (since ω≠0\omega \neq 0): A2−x2=ωx\sqrt{A^2 - x^2} = \omega x

Squaring both sides: A2−x2=ω2x2A^2 - x^2 = \omega^2 x^2

Substituting the values A=3A = 3 and x=2x = 2: 32−22=ω2(22)3^2 - 2^2 = \omega^2 (2^2) 9−4=4ω29 - 4 = 4\omega^2 5=4ω2  ⟹  ω2=54  ⟹  ω=52 rad/s5 = 4\omega^2 \implies \omega^2 = \frac{5}{4} \implies \omega = \frac{\sqrt{5}}{2} \text{ rad/s}

The time period TT is given by: T=2πω=2π5/2=4π5 sT = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{5}/2} = \frac{4\pi}{\sqrt{5}} \text{ s}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationsparticleexecuteslinearsimpleharmonic

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