A massless spring with spring constant 19 N/m hangs vertically. A body of mass 0.20 kg is attached to its free end and then released. Assume that the spring was un-stretched before the body was released. Find
a) How far below the initial position the body descends, and the
b) Frequency of the resulting Simple Harmonic Motion.
c) Amplitude of the resulting Simple Harmonic Motion.
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Answer:
a) At the point of release, GPE = 0, KE = 0, EPE of spring = 0
At max distance x below initial position,
loss in GPE = gain in EPE of spring (KE becomes 0 at that point)
mgx = ½kx²
mg = ½kx
(0.20)(9.81) = ½(19)(x)
x = 0.2065 m
x = 0.207 m (3 s.f.)
b)
= 1.55 Hz
c) Highest point of oscillation = initial position
Lowest point of oscillation = 0.2065 m below initial position.
Thus, amplitude of oscillation = 0.2065/2 = 0.103 m (3 s.f.)
