Question

A spring stretches 0.150m when a 0.300kg mass is gently suspended from it. The spring is then set up horizontally with the 0.300kg mass resting on a frictionless table.The mass is pulled so that the spring is stretched 0.100m from the equilibrium point and raised from rest. determine the stiffness constant and amplitude of horizontal oscillation

Answer 1

Spring tension under gravity weight:

x_1=0.15m

Weight in the vertical position of the spring:

m_1=0.3kg

Spring deviation from rest:

x_2=0.1m

Weight in the horizontal position of the spring

m_2=0.3kg

Solution:

According to Hooke's law, spring stiffness:

k=F/x_1 =mg/x_1 =0.3*9.81/0.15=19.6 N/m

Since the amplitude of oscillation cannot exceed the value of the level of displacement of the spring load from the equilibrium point, then:

A=x_2=0.1m

Answer:

Spring stiffness:

19.6 N/m

Amplitude of oscillation:

0.1m