Simple Harmonic Motion - A 75.0 N / m spring is suspended vertically with nothin

Question

Simple Harmonic Motion - A 75.0 N / m spring is suspended vertically with nothing attched to it and rests at equilibrium. Then a mass is attached to the spring and given a nudge. As a result, the mass oscillates between 11 cm and 17 cm below the original position of the unweighted spring.

Part A - Determine the mass attacthed to the spring and the period of the oscillation.

Part B - Determine the mass's maximum velocity. At what location will this maximum velocity occur?

Hint: What are the equilibrium position and the amplitude of the oscillation? (Please show steps! Thanks!)

Explanation / Answer

A)
Equilibrium position is at x = 14 cm = 0.14 m
At equilibrium:
M*g = K*x
M*9.8 = 75*0.14
M = 1.07 Kg
Answer: mass = 1.07 Kg

T= 2*pi*sqrt (M/K)
= 2*pi*sqrt (1.07/75)
= 0.75 s
Answer: period = 0.75 s

B)
w = sqrt (K/M) = sqrt (75/1.07) = 8.37 rad/s
A = (17-11)/2 = 3 cm =0.03 m

Max V = A*w = 0.03 * 8.37 = 0.25 m/s
Answer:
Maximum velocity = 0.25 m/s
It occurs at equilibrium position that is at x = 14 cm

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