Question

An object moving in simple harmonic motion has an amplitude of 0. 07 m and a maximum acceleration of 50 m/s2. What is the frequency of the system?.

Answer 1

Approximately
`42\; {\rm Hz}`.

Step-by-step explanation:

Let
`A` denote the amplitude and let
`f` denote the frequency of this Simple Harmonic Motion (SHM.) Assume that displacement is
`0` at time
`t = 0`.

The displacement of this SHM oscillator at time
`t` can be modelled with the sine function:

`x(t) = A\, \sin((2\, \pi\, f)\, t)`.

Apply the chain rule to differentiate displacement
`x(t)` with respect to time to find velocity
`v(t)`:

`v(t) = (2\, \pi\, f)\, A\, \cos((2\, \pi\, f)\, t)`.

Differentiate velocity
`v(t)` with respect to time to find acceleration
`a(t)`:

`a(t) = -(2\, \pi\, f)^(2)\, A\, \sin((2\, \pi\, f)\, t)`.

Note that as long as
`x(t) \\e 0`:

`\begin{aligned}(a(t))/(x(t)) &= (-(2\, \pi\, f)^(2)\, A\, \sin((2\, \pi\, f)\, t))/(A\, \sin((2\, \pi\, f)\, t)) = -(2\, \pi\, f)^(2)\end{aligned}`.

In other words, as long as
`x(t) \\e 0`, the ratio between
`x` and
`a` would be equal to
`(-1)\, (2\, \pi\, f)^(2)`.

The amplitude
`A` of an SHM is the maximum value of displacement
`x`. Additionally, the magnitude of acceleration
`a` is maximized whenever displacement
`x` is maximized.

In other words, when displacement is maximized,
`x = A = 0.07\; {\rm m}`. The magnitude of acceleration at that moment would also be maximized:
`|a| = 50\; {\rm m\cdot s^(-2)}`.

Since
`x > 0` and the direction of acceleration is opposite to that of displacement,
`a < 0`. Therefore,
`a = (-50)\; {\rm m\cdot s^(-2)}`.

Since
`x \\e 0`,
`(-1)\, (2\, \pi\, f)^(2) = a / x`. Therefore:

`\begin{aligned}f &= (1)/(2\, \pi)\, \sqrt{-(a)/(x)} \\ &= (1)/(2\, \pi)\, \sqrt{-((-50))/(0.07)}\; {\rm s^(-1)} \\ &\approx 42\; {\rm Hz}\end{aligned}`.