A mass m = 0.1 kg is hanging from a spring of unknown spring constant k. The mas

Question

A mass m = 0.1 kg is hanging from a spring of unknown spring constant k. The mass’s position y is measured from the attachment point of the spring, as shown. You observe the position vs time curve below.

A. What are the period T and amplitude A of the oscillation? What is the equilibrium length of the spring (with the hanging mass attached)? IPLS

B. Write down a function y(t) that describes the position of the mass. Think carefully about the phase shift!

C. What is the spring constant k? As always, include appropriate units.

D. If you doubled k, what would happen to the period T? If you doubled k and doubled m, what would happen to the period?

E. What is the unstretched length of the spring?

Explanation / Answer

Lets first find the period of the oscillations, then we can obtain an equation for the motion. The period T = 2 q m/k. The mass m is 0.1 Kg. To find k, we use the fact that 100 grams causes the spring to stretch an additional 10 cm.

C) F = kx,

we have mg = kx

0.1(9.8) = k(0.1) k = 9.8 N/m

A&B) The period of the motion is therefore T = 2 q 0.1/9.8 0.635 sec

At t = 0 the mass is at its maximum distance from the origin. Thus, y(t) = 0.6 cos(2t/T). Using T = 0.635 sec gives y(t) = 0.6 cos(2t/0.635) 0.6 cos(9.9t) The cosine function is the appropriate one, since at t = 0 the mass is at its maximum distance from equilibrium.

D) if doubled the K value, Time period will be decreses to 0.202 sec, if i doubled both k and m decreases to 0.0640 sec

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