Question

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 8 inches above the equilibrium position.

Find the equation of motion. (Use g = 32 ft/s2 for the acceleration due to gravity.)

x(t)= ?

Answer 1

`x(t) = 8\cdot \cos 9.798\cdot t`

Explanation:

The equation of motion for the spring-mass system is:

`x(t) = A\cdot \cos (\omega\cdot t + \phi)`

Where:

`\omega = \sqrt{(k)/(m)}`

`k` - Spring constant, in
`(lbm)/(s^(2))`.

`m` - Mass, in
`lbm`.

The spring constant is:

`k = (m\cdot g)/(\Delta s)`

`k = ((24\,lbm)\cdot (32\,(ft)/(s^(2)) ))/((4)/(12) \,ft)`

`k = 2304\,(lbm)/(s^(2))`

The angular frequency is:

`\omega = \sqrt{(2304\,(lbm)/(s^(2)) )/(24\,lbm) }`

`\omega \approx 9.798\,(rad)/(s)`

The initial condition for the system is:

`x(0) = +8\,in`

`v(0) = 0\,(in)/(s)`

The function for speed is obtained by deriving the previous function:

`v(t) = -\omega \cdot A\cdot \sin (\omega\cdot t + \phi)`

The following expressions are formed by substituting all known variables:

`A \cdot \cos \phi = 8\,in`

`-(9.798\,(rad)/(s) )\cdot A \cdot \sin \phi = 0\,(in)/(s)`

The phase angle is found by dividing the initial velocity by the initial position:

`\tan \phi = 0`

`\phi = 0\,rad`

The amplitude is:

`A = 8\,in`

The equation of motion is:

`x(t) = 8\cdot \cos 9.798\cdot t`