1. sqrt5/2π 2. 4π/sqrt5 3. 2π/sqrt3 4. sqrt5/π

Answer: 2. 4π/sqrt5 Key idea: in SHM v=ω√(A²-x²) and a=ω²x . At x=2 cm, A=3 cm: ω√(9-4)=ω²×2⇒ω=sqrt5/2 s ⁻¹ . T=2π/ω=4π/sqrt5 s.

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 :

1. \(\frac{\sqrt5}{2\pi}\)

2. \(\frac{4\pi}{\sqrt5}\)

3. \(\frac{2\pi}{\sqrt3}\)

4. \(\frac{\sqrt5}{\pi}\)

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

Best answer

Answer: 2. \(\frac{4\pi}{\sqrt5}\)

Key idea: in SHM \(v=\omega\sqrt{A^2-x^2}\) and \(a=\omega^2x\).

At \(x=2\) cm, \(A=3\) cm: \(\omega\sqrt{9-4}=\omega^2\times2\Rightarrow\omega=\dfrac{\sqrt5}{2}\) s\(^{-1}\).

\(T=\dfrac{2\pi}{\omega}=\dfrac{4\pi}{\sqrt5}\) s.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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