A 75 N/m spring is suspended vertically with nothing attached to it and rests at

Question

A 75 N/m spring is suspended vertically with nothing attached 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. Find the equilibrium position and the amplitude of the oscillation to help.

a. Determine the mass attached to the spring and the period of the oscillation?

b. Determine the mass's maximum velocity. At what lovation will this maximum velocity occur?

Explanation / Answer

a) equilibrium would be average position = (17+11)/2 = 14 cm

amplitude = 3 cm

so kx = mg to amplitude

75*.14 = m*9.81

m=1.07 kg

T = 2 pi sqrt(m/k) = 2*pi*sqrt(1.07/75)=0.75 s

b) 1/2 k a^2 = 1/2 mv^2

v = sqrt(75/1.07)*0.03=0.251 m/s

will happen at equilibrium so 14 cm

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