A 1.07 kg mass is hung from a spring as shown in the diagram. The mass is pulled
ID: 1572265 • Letter: A
Question
A 1.07 kg mass is hung from a spring as shown in the diagram. The mass is pulled down a short distance and released. A stopwatch is used to measure the time of 17 oscillations. It is found that these 17 oscillations take t = 13.2 s. What is the spring constant of this spring? An additional mass of 0.997 kg is added to the spring. What is the frequency of the spring now? When the additional mass was added by how much did the equilibrium length of the spring change? Assume the spring is on the surface of the Earth.Explanation / Answer
frequency = 17/13.2 = 1.288 Hz
f= 1/2pi * sqrt(k/m)
1.288 = 1/2pi* sqrt(k/1.07)
k = 70.06 N/m
f(new) = 1/2pi* sqrt(70.06 / 1.07 + 0.997)
= 0.93 Hz
extension earlier = mg / k = 1.07*9.8 / 70.06 = 0.1496 m
extension after adding the mass = (1.07+0.997) *9.8 / 70.06 = 0.2891 m
extension = 0.2891 - 0.1496 = 0.1395 m = 13.95 cm
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