The phase of a particle performing a linear S.H.M increases by (π^c)/(6) after every 5 second. The frequency of its oscillation is

Correct answer: D. (1)/(60)Hz The phase of a particle in S.H.M is given by φ = ω t + φ_0 The change in phase over a time interval Δ t is Δ φ = ω Δ t Given Δ φ = (π)/(6) and Δ t = 5 s ω × 5 = (π)/(6) ω = (π)/(30) rad/s The frequency of oscillation f is given by f = (ω)/(2π) f = (π/30)/(2π) = (1)/(60) Hz Answer: (1)/(60)Hz

MHT CET 202622 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 22 April
Shift: Morning Shift
Subject: Physics
Chapter: Oscillations

MHT CET 2026 · 22 April · Morning Shift · Physics · Oscillations

The phase of a particle performing a linear S.H.M increases by...

The phase of a particle performing a linear S.H.M increases by

\dfrac{\pi^c}{6} after every 5 second. The frequency of its oscillation is

Options

A $1\,\text{Hz}$
B $\dfrac{1}{20}\text{Hz}$
C $\dfrac{1}{30}\text{Hz}$
D $\dfrac{1}{60}\text{Hz}$

Correct answer

D. $\dfrac{1}{60}\text{Hz}$

Step-by-step solution

The phase of a particle in S.H.M is given by

\phi = \omega t + \phi_0

The change in phase over a time interval

\Delta t

\Delta \phi = \omega \Delta t

Given

\Delta \phi = \dfrac{\pi}{6}

and

\Delta t = 5 \text{ s}

\omega \times 5 = \dfrac{\pi}{6}

\omega = \dfrac{\pi}{30} \text{ rad/s}

The frequency of oscillation

f

is given by

f = \dfrac{\omega}{2\pi}

f = \dfrac{\pi/30}{2\pi} = \dfrac{1}{60} \text{ Hz}

Answer:

\dfrac{1}{60}\text{Hz}

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