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 distan…

Question

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 [NEET 2018]

Accepted Answer

Correct answer: B. (b) The acceleration of particle executing SHM at any instant (at position x) is given as a =- ^2x ⇒ |a |= ^2x …(i) Here, x=5 m , |a |=20 ms^-2 On substituting the given values in Eq. (i), we get 20= ^2 5 ⇒ ^2= 205=4 or =2 rad s^-1 As we know that, time period, T= 2 …(ii) On substituting the value of in Eq. (ii), we get T= 2 2= s

$A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m / s^{2} at a distance of 5 m from the mean position. The time period of oscillation is [NEET 2018]$

Solution

More Simple Harmonic Motion PYQs