Question

A spring, of negligible mass and which obeys Hooke's Law, supports a mass M on an incline which has negligible friction. The figure below shows the system with mass M in its equilibrium position. The spring is attached to a fixed support at P. The spring in its relaxed state is also illustrated. Mass M has a value of 255 g. Calculate k, the spring constant. The mass oscillates when given a small displacement from its equilibrium position along the incline. Calculate the period of oscillation.

Answer 1

Length of stretched spring

from figure change of x length = (80-10) = 70 cm

change of length y length = (70-20) = 50 cm

Length L = sqrt( 0.7^2 + 0.5^2) = 0.86 m

from figure agian Length of unstretched spring = 0.7 -0.2 = 0.5 m

Angle of incline = tan theta = 0.5/0.7 = 0.714

theta = 35.53 deg

So

spring costant k = mg sin theta /(change of length

K = 0.255 * 9.81 * sin 35.53/(0.86 - 0.5)

K = 4.038 N/m

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time period T = 2pi sqr(m/k)

T = 2*3.14 * sqrt( 0.255/4.038)

T = 1.57 secs