A helical spring, of negligible mass, and which is found to extend 0.25 mm under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is displaced vertically through 12.5 mm and released. Determine the frequency of natural vibration of the system.

A helical spring, of negligible mass, and which is found to extend 0.25 mm under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is displaced vertically through 12.5 mm and released. Determine the frequency of natural vibration of the system. Correct Answer 4.98 Hz

Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend the spring by an amount, δ = 0.25/1.5 x 60 = 10 mm = 0.01m We know that frequency of the system, n = 1/2π√g/δ = 1/2π√9.81/0.01 = 4.98 Hz.

Related Questions

Statement: Resonance is a special case of forced vibration in which the nature and frequency of vibration of the body are the same as the impressed frequency and the amplitude of forced vibration, is maximum. Reason: The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force.