Not So Simple Harmonic Motion

[Timcago]

A mass suspended from a spring oscillates in simple harmonic motion. the mass completes two cycles every second. The distance between the highest and lowest points in the oscillation is 10cm. When the mass is at rest, the spring is 22cm long. Construct a function modeling the length of the spring after t seconds when the mass starts at the maximum height.

ANyone know how to solve this one?

 

[soroban]

Hello, Timcago!

A mass suspended from a spring oscillates in simple harmonic motion.
The mass completes two cycles every second.
The distance between the highest and lowest points in the oscillation is 10cm.
When the mass is at rest, the spring is 22cm long.
Construct a function modeling the length of the spring after t seconds
when the mass starts at the maximum height.

-     *
:     |
:     |
:     |
:     |
:     |
22    |
:   - |             * *
:   : |           *     *
:   5 |          *       *
:   : |
-   - * - - - - * - - - - * - - - - - - -
    : |                  ½ sec
    5 |*       *
    : | *     *
    - |   * *

We know that the amplitude is 5: \(\displaystyle \:y\;=\;5\cdot\sin(kt)\)

It complete a cycle in a half second.

\(\displaystyle \;\;\)Hence: \(\displaystyle \:y\;=\;5\cdot\sin(4\pi t)\)


Therefore, the length of the spring is: \(\displaystyle \:L\;=\;22\,+\,5\cdot\sin(4\pi t)\)

 

[Timcago]

Wouldnt it be cos since the mas starts at the maximum height?

I thought sin was only when it started from its baseline.

 

[skeeter]

Wouldnt it be cos since the mas starts at the maximum height?

I thought sin was only when it started from its baseline.

You did not provide that information in your original post ... also, you did not state where the value for zero height of the mass is to be established.

in simple harmonic motion problems, zero is usually where the mass is in a state of equilibrium. if it starts at its maximum height, then ...

\(\displaystyle h = 5cos(4\pi t)\)