A lvl H2 Phy: Simple Harmonic Motion

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.)

A lvl H2 Phy: Simple Harmonic Motion

A mass on the end of a light helical spring is given a vertical displacement of 3.0 cm from its rest position and then released. If the subsequent motion is simple harmonic with a period of 2.0 s. through what distance will the bob move in the first 0.75 sec?

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Answer:

A = 3
ω = 2 π / T = 2 π / 2 = π

x = A cos (ωt) since when x = A when t = 0

At t = 0.75, x= 3 cos (0.75 π) = -2.1 cm

Since motion has not reached the other end of the amplitude (which is -3 cm and at t = 1 s),
total distance travelled = 3 cm + 2.1 cm = 5.1 cm