A horizontal rod of mass m = 3 k π 2 k g and length L is pivoted at one end. The rod at the other end is supported by a spring of force constant k . The rod is displaced by a small angle θ from its horizontal equilibrium position and released. The time period (in second) of the subsequent simple harmonic motion is

Question

A horizontal rod of mass m = 3 k π 2 k g and length L is pivoted at one end. The rod at the other end is supported by a spring of force constant k . The rod is displaced by a small angle θ from its horizontal equilibrium position and released. The time period (in second) of the subsequent simple harmonic motion is,

MARKS App · Accepted Answer

E= 1 2 m L 2 3 ω 2 + 1 2 k x - x 0 2 + m g x 2 d E d t = 1 2 m L 2 3 2 v L a L + k x - x 0 v + m g 2 v = 0 = m 3 a + k x - k x 0 + m g v = 0 At equilibrium, k x 0 = m g 2 ∴ m a 3 = - k x a = - 3 k m x T = 2 π ω = 2 π m 3 k = 2 π 3 k 3 k π 2 = 2 s

Search any question & find its solution

Looking for more such questions to practice?