Question

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose it is displaced 0.122 m from its equilibrium position and released with zero initial speed. After 0.810 s, its displacement is found to be 0.122 m on the opposite side and it has passed the equilibrium position once during this interval. Find:
a. The amplitude
b. The period
c. The frequency.

Answer 1

Answer:

a. Amplitude = 0.244 meters

b. Period = 1.62 seconds

c. Frequency = 0.6173 Hz

Explanation:

a.

With the position goes from 0.122 meters to -0.122 meters (negative because it is in the opposite side of the equilibrium point), the amplitude is the maximum value minus the minimum value:

Amplitude = 0.122 - (-0.122) = 0.122 + 0.122 = 0.244 meters

b.

The period is the amount of time the object takes to arrive in the same position again. So, if it takes 0.81 seconds to go to -0.122 m, it will take another 0.81 seconds to come back to 0.122 m, so the period is the sum of these two times:

Period = 0.81 + 0.81 + 1.62 seconds

c.

The frequency of the movement is the inverse of the period:

Frequency  = 1 / Period

So if the period is 1.62 seconds, the frequency is:

Frequency = 1 / 1.62 = 0.6173 Hertz