A 2 kg mass rests on a horizontal surface. The mass is attached to a massless sp
ID: 1997619 • Letter: A
Question
A 2 kg mass rests on a horizontal surface. The mass is attached to a massless spring with force constant 600 N/m. The spring is anchored to a vertical wall. The mass is displaced such that the spring is extended by 8 cm from its equilibrium position. After 5 complete cycles, the mass is observed to be displaced 2 cm from the equilibrium position. (a) How much energy has been lost from the system, and where are some of the places it may have gone? (b) Assuming the damping force is linearly proportional to the velocity, what is the period of the oscillator?
Explanation / Answer
(A) initial Spring PE energy = k A^2 / 2
= 600 x (0.08^2) / 2 = 1.92 J
initial gravitational PE = 0
final spring PE = 600 (0.02^2) / 2 = 0.12 J
final gravt PE = 2 x 9.8 x (0.08 - 0.02) = 1.176 J
energy lost = (1.92 ) - (0.12 + 1.176) = 0.624 J
It may have lost in heat
(B) A = Ao e^(-bt/2m)
0.02 = 0.08 e^(-5bT / 2m)
5 b T /2m = 1.386
b/2m = 0.277/ T
and w = 2pi / T = sqrt[ (k/m) - (b/2m)^2 ]
4 pi^2 / T^2 = ( 600/2) - (0.277 / T)^2
157.99 / T^2 = 300
T = 0.726 sec
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