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.

Question

drwls · Accepted Answer

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

Mlungisi · Answer

(a) 2,83ms (b) 2,11ms (c) 2,00j (d) x=acoswt .