- #1

VincentweZu

- 13

- 0

## Homework Statement

A mass

*m*is attached to a horizontal spring of spring constant

*k*. The spring oscillates in simple harmonic motion with amplitude

*A*. Answer the following in terms of

*A*.

At what displacement from equilibrium is the speed half of the maximum value?

## Homework Equations

PE

_{1}+ KE

_{1}= PE

_{2}+ KE

_{2}

## The Attempt at a Solution

[itex]\frac{1}{2}[/itex]kA

^{2}= [itex]\frac{1}{2}[/itex]mv

_{max}

^{2}

kA

^{2}= mv

_{max}

^{2}

kA

^{2}/m = v

_{max}

^{2}

[itex]\frac{1}{2}[/itex]kA

^{2}= [itex]\frac{1}{2}[/itex]mv

^{2}+ [itex]\frac{1}{2}[/itex]kΔx

^{2}let v = v

_{max}/2

kA

^{2}= m(v

_{max}/2)

^{2}+ kΔx

^{2}

kA

^{2}= m(v

_{max}

^{2})/4 + kΔx

^{2}

kA

^{2}= m(kA

^{2}/m)/4 + kΔx

^{2}

kA

^{2}= kA

^{2}/4 + kΔx

^{2}

A

^{2}= A

^{2}/4 + Δx

^{2}

(3/4)A

^{2}= Δx

^{2}

Δx = ([itex]\sqrt{3}[/itex]/2)A

the answer that I arrived at was ([itex]\sqrt{3}[/itex]/2)A, however the answer is 1/2. If I subbed in arbitrary numbers for A, m, and k and let Δx = ([itex]\sqrt{3}[/itex]/2)A then the velocity that results is half the max velocity.

I would like someone to confirm my answer or find something wrong with it so that I know which is the right answer. Thanks.