A spring is suspended from a support and has a 0.060kg weight pan attached to it

Question

A spring is suspended from a support and has a 0.060kg weight pan attached to it. The bottom edge of the weight pan is at the 60gm mark of an adjacent vertical meter stick. You place a 0.050kg mass onto the weight and gently lower the mass so the spring hangs motionless and observe that now the bottom of the weight pan is at the 72cm mark. a. What is the spring constant? b. Neglecting the mass of the spring, if you pull the mass down a bit and let it oscillate, what will be the frequency in Hz? (total mass in 0.110kg)

Explanation / Answer

Part A:

mg = 1/2*kx2

0.05*9.8 = 0.5*k*[(72 - 60)*10-2]2

Sping constant, k = 68.0555 N/m2

Part B:

Frequency, f = {1/[2*pi]}*[k/m]1/2 = 0.1592*[68.0555/0.11]1/2 = 3.9587 Hz

Frequency = 3.9587 Hz OR 4 Hz

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