Answer: SHM: 45g Mass, .35m Amp - Speed Half Max Velocity

[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?

 

[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.

 

[wais]

No, you are not wrong. The correct answer is indeed A. .3 m. The equation you used is correct, and by solving for x, you get the correct distance from equilibrium for the mass to have a speed that is half its maximum velocity. Good job!