# Simple harmonic motion (finding omega)

## Homework Statement

A particle executes simple harmonic motion with an amplitude of 2.51cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its max speed?

## Homework Equations

A=2.51 cm
x(t) = Acos(wt)
v(t) = -Awsin(wt)
vmax = |-Aw|

## The Attempt at a Solution

I just want to know how to find w so I can plug it into the vmax equation.

## Homework Statement

A particle executes simple harmonic motion with an amplitude of 2.51cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its max speed?

## Homework Equations

A=2.51 cm
x(t) = Acos(wt)
v(t) = -Awsin(wt)
vmax = |-Aw|

## The Attempt at a Solution

I just want to know how to find w so I can plug it into the vmax equation.

You actually don't need to plug it in. If $$v_{max} = Aw$$ then $$\frac{v_{max}}{2} = \frac{Aw}{2}$$

Just use the energy equation

$$\frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}mv^2$$

where $$v= \frac{Aw}{2}$$, and remember to sub $$\frac{k}{m}$$ for $$w^2$$.

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You actually don't need to plug it in. If $$v_{max} = Aw$$ then $$\frac{v_{max}}{2} = \frac{Aw}{2}$$

Just use the energy equation

$$\frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}mv^2$$

where $$v= \frac{Aw}{2}$$, and remember to sub $$\frac{k}{m}$$ for $$w^2$$.

I don't know what k is though. How do I find k?

Divide both sides by k!

O ok haha sorry I'm really tired. Thanks for the help I solved it!

You don't need the velocity or the angular frequency in order to answer the question. You're only asked at what displacement the velocity has half its maximum value.

$$x=A\cos{(\omega t)}$$

$$v=-\omega A \sin{(\omega t)}$$

$$v=-\omega A \sqrt{(1-\cos ^2 {(\omega t))}}$$

$$v = -\omega \sqrt{(A^2-x^2)}$$

$$|v|=\tfrac{1}{2} v_{max}=\tfrac{1}{2}\omega A$$

$$\tfrac{1}{2}\omega A = \omega \sqrt{(A^2-x^2)}$$

You don't always need all the constants involved in a problem in order to solve for a particular value. :)