Question

A mass hanging from a spring oscillates with a period of 0.35 s.

Suppose the mass and spring are swung in a horizontal circle, with

the free end of the spring at the pivot. What rotation frequency, in

rpm, will cause the spring’s length to stretch by 15%?

Answer 1

Mass of the object attached to the spring = m

Spring constant = k

Length of the spring = L

Time period of oscillation = T = 0.35 sec

Time period of oscillation of a simple harmonic motion is given by,

m = (3.103x10^-3)k

Now the mass is swung around in a horizontal circle such that the spring stretches by 15%.

Extension of the spring = X = 0.15L

Radius of rotation of the object = R = L + X = L + 0.15L = 1.15L

Angular velocity of the object =

The centripetal force necessary for the circular motion of the object is provided by the spring.

m²R = kX

(3.103x10^-3)k²(1.15L) = k(0.15L)

= 6.483 rad/s

Converting to rpm,

= 6.483 x (60/2) rpm

= 61.91 rpm

Rotation frequency = 61.91 rpm