# Vibrations and Waves

1. Jan 3, 2005

### Soaring Crane

How far from equilibrium will a 45 g mass with amp. of .35 m for a spring (k = 21 N/m) undergoing simple harmonic motion have a speed that is half its maximum velocity?
a. .3 m
b. .45 m
c. .7 m
d. .92 m
e. 3.5 m

Choice B is wrong.

Use

v = v_max*sqrt[1 - (x^2/A^2 m)]

Do I use k or m at all???
(1/2)v_max = v_max*sqrt[1 - (x^2/A^2 m)]
Left with:

1/2 =sqrt[1 - (x^2/A^2 m)]

After plugging everything in and solving for x I get .303m (A.). Am I wrong?

Last edited: Jan 3, 2005
2. Jan 3, 2005

### dextercioby

Okay,let's take it this way.'A' is the only possible answer as it has the only figure less than the amplitude of oscillation.

Another rigurous mathematical justification would be:
$$x=A\sin\omega t$$(1)
$$v=\omega A\cos\omega t$$(2)
From (2) u get that for hamf of the maximum velocity the 'cos' must be 1/2,therefore the 'sin' would be $\sqrt{3}/2$.
Plug the sine into (1) and the "x" comes out to be $0.175\sqrt{3}m$
,which can be (very roughly) approximated to 0.30m.

Daniel.