Posted by **Amber** on Friday, May 28, 2010 at 10:11pm.

A 0.50-kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.15 m. Determine (a) the velocity when it passes the equilibrium point, (b) the velocity when it is 0.10 m from equilibrium, (c) the total energy of the system, and (d) the equation describing the motion of the mass, assuming that at t = 0, x was a maximum.

- Physics -
**drwls**, Friday, May 28, 2010 at 10:42pm
Here some facts about simple harmonic motion to help you answer these yourself.

(a) The angular frequency is w = 2 pi f radians per second.

Tha maximum velocity is w A, where a is the amplitude.

The spring constant k is related to w and M by

w = sqrt(k/m)

Compute w. You will need it later.

(b) 0.10 m is 2/3 of the maximum deflection. Potential energy is proportion do the square of deflection, and will be (2/3)^2 = 4/9 of the maximum energy, or

(4/9)*(1/2)*k*(Amplitude)^2. The velocity will be (5/9) of the maxiomum energy. Use that to solve for velocity.

(c) I already mentioned that the toal energy is (1/2)*k*(Amplitude)^2

(d) X = A cos wt

- Physics -
**Mlungisi**, Thursday, June 16, 2011 at 7:12am
(a) 2,83ms (b) 2,11ms (c) 2,00j (d) x=acoswt .

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