A spring, whose unstretched length is 12 cm, is suspended from the ceiling and a

Question

A spring, whose unstretched length is 12 cm, is suspended from the ceiling and a 1.5 kg object is attached to it. The object is released from an unknown height and oscillates vertically about an equilibrium position 28 cm below the ceiling. It passes through this equilibrium position with a speed of 8 cm/s. a) What is the period of the motion? b) What is the maximum stretch the spring has from its 12 cm length? Something to consider here... since it is a vertical situation, the added mass causes the spring to stretch from its natural length. The added mass just 'hangs there', at the new equilibrium, with the force of gravity balanced by the spring force upwards (think Newton's laws). Knowing this, you need to get .kof the spring. Only when you further displace the mass away from this 'hanging there' location will it oscillate AROUND THIS NEW POINT with a unique amplitude (depending on how much it is displaced from this new equilibrium)

Explanation / Answer

(A) vmax = 8cm / s

A = 28 - 12 = 16 cm

v_max = A w

w = (8 cm/s) / 16 = 0.5 rad/s

and T = 2pi/ w = 4 pi sec or 12.6 sec

(B) maximum stretched length, L = 28 + 16 = 44 cm

L0 = 12 cm

max stretch = L - L0 = 32 cm

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