Question

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 250 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.

(a) Calculate the amplitude of the motion.
_______________________________ m
(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]
_______________________________ m/s

Answer 1

m = 3.5 kg
x = 0.020 m
v = 0.55 m/s
k = 250 N/m
since w = sqrt(k/m) = sqrt (250/2) = 11.18 sec ^ -1
and
v = w sqrt ( A^2 - x^2)
where A is the Amplitude and x is displacement at any time
so

0.55 / 11.18 = sqrt ( A^2 - x^2)
0.04919 = sqrt ( A^2 - x^2)
0.002420 = A^2 - (0.020)^2
0.002420 + 0.0004 = A^2
0.0531 = A
a ) so the Amplitude is 0.0531 m.
b) v(max) = wA
so v(max) = 11.18 * 0.0531
v(max) = 0.5937m/s