Question

A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were attached
to the spring, at what angular frequency (in rad/s) would it oscillate?

Answer 1

Answer:

Approximately $3.14\; {\rm rad \cdot s^{-1}}$.

Explanation:

Fact: the angular velocity $\omega$ of a simple harmonic oscillator is the ratio between the maximum velocity $v_{\text{max}}$ and the maximum displacement $x_\text{max}$ of this oscillator. In other words:

$\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}$.

Derivation of the previous equation:

Let $A$ denote the amplitude of this oscillation, and let $\omega$ denote the angular velocity.

The displacement of the oscillator at time $t$ would be:

$x(t) = A\, \sin(\omega\, t)$.

The maximum displacement of this oscillator would be $x_\text{max} = A$.

The velocity of this oscillator at time $t$ is the derivative of displacement with respect to time:

$\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}$.

The maximum velocity of this oscillator would be $v_\text{max} = A\, \omega$.

Notice that dividing $v_\text{max} = A\, \omega$ by $x_\text{max} = A$ would give:

$\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega$.

It is given that $v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}$ while $x_\text{max} = 0.0880\; {\rm m}$. Therefore:

$\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}$.

(Radians per second.)