In a lab, a student measures the unstretched length of a spring as 12.1 cm. When

Question

In a lab, a student measures the unstretched length of a spring as 12.1 cm. When a 125.8-g mass is hung from the spring, its length is 21.1 cm. The mass-spring system is set into oscillatory motion, and the student observes that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles.

a) Calculate the period of oscillation for this system, assuming no damping. (solved, T = 0.602 s)
b) What would be the difference between the period with no damping and the period with damping?

Explanation / Answer

mass m = 8.60 kg spring constant, k =400N/m At equilibrium position , Kx = mg extension x = mg / K =0.210m from this original length of the spring L= 0.370 - x =0.16 m new mass M = 12.40 kg At equilibrium position , Kx ' = Mg extension x ' = Mg / K =0.303 ? new total length of the spring = L + x ' =0.463 m

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