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Oscillations question

2018 · Q156
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Oscillations question

2018 · Q156

NEETPhysicsOscillationsMCQ+4 / −1
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s–2 at a distance of 5 m from the mean position. The time period of oscillation is
  1. A
    2π\piπ s
  2. B
    π\piπ s
  3. C
    2 s
  4. D
    1 s
View written solutionFree

Correct answer: B

Given, acceleration, a = 20 m/s2,
and displacement, y = 5m

Magnitude of acceleration of a particle
moving in a SHM is, |a| = ω2{\omega ^2}ω2y; where y is amplitude.

⇒\Rightarrow⇒ 20 = ω2{\omega ^2}ω2(5)

⇒\Rightarrow⇒ ω=2\omega = 2ω=2 rad/s

Time period of pendulum,

T = 2πω=2π2{{2\pi } \over \omega } = {{2\pi } \over 2}ω2π​=22π​ = π\pi π s

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