A block of mass 0.125 kg is hanging on a spring. Nora gently pulls the mass down

Question

A block of mass 0.125 kg is hanging on a spring. Nora gently pulls the mass down a distance of 4.0 cm and then lets go. The mass bobs up and down in simple harmonic motion (i.e. it oscillates) with a period of 0.47 s.

(a) What is the value of the spring constant? N/m

Nora stops the mass from oscillating. She gently pulls the mass down again, this time to a distance of 5 cm, and lets the mass bob up and down.
(b) What is the period of oscillation? s

Nora again stops the mass from oscillating. She takes the mass off of the spring and notes the position of the bottom of the hanging spring. She gently hangs the mass back onto the spring, this time making sure it does NOT oscillate.

By what distance did the spring stretch? In other words, what is the displacement of the spring with the mass hanging from it?
HINT: Draw a free body diagram that shows the two forces acting on the mass when it is hanging and not oscillating and remember Hooke's Law (and our written assignment). Note that your answer should be in cm.
cm

Explanation / Answer

Given

mass of the block m = 0.125 kg

the displacement is x = 0.04 m

time period is T = 0.47 s

a)

we know that the time period T = 2pi sqrt(m/k)

T^2 = 4 pi^2(m/k)

k = 4 pi^2(m/T^2)

k = 4 pi^2 (0.125/0.47^2) N/m

k = 22.34 N/m

b) the time period does not depend on the displacement so the same timeperiod for the displacement of x = 0.05 m also

if the spring does not making oscillations means the forces acting on the mass are  

F = -kx = mg

x = mg/k

x = 1.125*9.8/(22.34) m

x = 0.49351 m

x = 49.351 cm

so the spring was stretched by 49.351 cm  

the forces acting on the block are

one is the elastic force upward F = kx and the other force gravitational pull acting downward if these are equal and acting in opposite directions then the block will be in equilibrium making zero oscillations

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