Question

A spring is hung from the ceiling. A 0.422 -kg block is then attached to the free end of the spring. When released from rest, the block drops 0.130 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

Answer 1

Mass of block = 0.422 kg

Max Stretch = Amplitude A = 0.130 m

Using energy conservation at equilibrium point and max stretch point

KEi + PEi = KEf + PEf

KEi = 0, Since released from rest

KEf = 0, at max stretch speed is zero

0 + m*g*h0 = 0 + 0.5*k*d^2 + m*g*hf

k*d^2 = 2*m*g*(h0 - hf)

Notice that d = (h0 - hf)

k = 2*m*g*(h0 - hf)/d^2

k = 2*0.422*9.81*0.130/0.130^2

k = 63.69 N/m

Part B

Now angular frequency is given by:

w = sqrt (k/m)

w = sqrt (63.69/0.422)

w = 12.28 rad/sec

Please Upvote.