A frictionless spring with a 10kg mass can be held stretched

A frictionless spring with a 10-kg mass can be held stretched 1.4 meters beyond its natural length by a force of 60 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2.5 m/sec, find the position of the mass after tseconds.

Solution

Answer:-   Below is an equation for position vs time x = A*sin(2**f*t)
A = amplitude, f = frequency, t = time
Frequency = 1/(2*)*(k/m)^0.5

Force = k*distance stretched
60 = k*1.4
k = 60 ÷ 1.4 = 42.86 N/m
Frequency = 1/(2*)*(42.86/10)^0.5=0.33 .

The initial Kinetic energy of the mass = (1/2)*10*2.5^2 =31.25 Joules
As the mass moves its kinetic energy decreases to 0 Joules as the potential energy of the spring increases to the maximum value.

Increase of PE = (1/2)*k*distance^2 = (1/2)*42.86*d^2
(1/2)*42.86*d^2 =31.25
d = 1.86m
This is the amplitude of the harmonic motion of the mass.
Amplitude =1.86meters

x = A*cos(2**f*t)
x = 1.86*sin(2**0.33*t)

x = 1.86*sin(2.07*t)