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
