We have a spring of negligible mass hanging from the ceiling, initially in its e

Question

We have a spring of negligible mass hanging from the ceiling, initially in its equilibrium position. Then we hang a mass from it and the mass and spring start oscillating. At the bottom of its trajectory the mass is 5 centimeters from equilibrium. (a) What's the period of oscillation? (b) What is the speed of the mass when it is 1.25 cm below its original equilibrium? (c) If we add a 12 gram mass to the mass that we added previously, the period of oscillation doubles. What was the mass that we originally added?

Explanation / Answer

a] mg = kA

m/k = A/g = 0.05/9.8 = 0.005102

T = 2 pi* sqrt(m/k)

= 2 pi* sqrt(0.005102)

= 0.4488 s

b) 0.5 mv^2 = 0.5kA^2 - 0.5kx^2

v^2 = (k/m) (A^2 - X^2)

v = sqrt[1/0.005102*(0.05^2-0.0125^2)]

= 0.67778 m/s

c] if period doubles, mass becomes 4 times.

4m = m +12 g

m = 4 gram answer

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