Question

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, the resulting undamped mass-spring system is disturbed from its rest state by the force F(t) = 170 cos(5t) and no initial velocity. The force F(t) is expressed in Newtons (am) and time is measured in seconds. What Initial Value Problem models this situation?

Answer 1

`15(d^(2)y(t))/(dt^(2) )  - 441.45y(t) =` ±
`170 cos(5t)`

`y(0)=0`,
`y'(0)=0`

Explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are:
`F_(r) (t)` that correspond to the force of resistance on the mass by the action of the spring and
`F(t)` that is an external force with unknown direction (that does not specify in the enounce).

For determinate
`F_(r) (t)` we can use Hooke's Law given by the formula
`F_(r) (t) = k y(t)` where
`k` correspond to the elastic constant of the spring and
`y(t)` correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

`k = (F_(r))/(y) = (mg)/(y) = (15 Kg (9.81 (m)/(s^(2) ) ))/((1)/(3) m)  = 441.45 (N)/(m)`

Now we apply Newton's 2nd Law and obtaint that...

`F_(r) (t)` ±
`F(t)` =
`ma(t)`

`F_(r) (t) = ky(t) = 441.45y(t)`

`F(t) = 170 cos(5t)`

`m = 15 kg`

`a(t) = (d^(2)y(t))/(dt^(2) )`

Finally...
`15(d^(2)y(t))/(dt^(2) )  - 441.45y(t) =` ±
`170 cos(5t)`

We know from the problem that there's not initial displacement and initial velocity, so...
`y(0)=0` and
`y'(0)=0`

Finally the Initial Value Problem that models the situation describe by the problem is

`\left \{ 15(d^(2)y(t))/(dt^(2) )  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.`