Question

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. Find the velocity of the mass, when it is 5 mm below its rest position.

(a) 0.36 m/s

(b) 0.46 m/s

(c) 0.56 m/s

(d) none of the mentioned

I got this question by my college professor while I was bunking the class.

This is a very interesting question from Centre of Percussion topic in section Simple Harmonic Motion of Machine Kinematics

Answer 1

Right answer is (a) 0.36 m/s

The explanation: 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

Let v = Linear velocity of the mass.

We know that angular velocity,

ω = √g/δ = √9.81/0.01 = 31.32 rad/s

and

v = ω√r^2 – x^2

  = 31.32√(0.0125)^2 − (0.005)^2 = 0.36 m/s.