Question

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. It is displaced a distance 0.120 mm from its equilibrium position and released with zero initial speed, then after a time 0.790 ss its displacement is found to be a distance 0.120 mm on the opposite side, and it has passed the equilibrium position once during this interval.Find the amplitude of the motion.
=________________m

Find the period of the motion.

=________________s

Find the frequency of the motion.

=_________________Hz

Answer 1

Answer:

a.0.120mm

b.1.58s

c.0.6329Hz

Explanation:

a. Given that 0.120mm is displaced from equilibrium, 0.120mm after 0.790s on opposite side:

-The amplitude is the maximum displacement from equilibrium.

-The object goes from x=+A to x=-A and back during one cycle.

#Hence, the amplitude of the motion is 0.120mm

b.Motion from maximum positive displacement to maximum negative displacement takes places during half the period of Simple Harmonic Motion(SHM)

$0.790s=T/2\\\\T=0.790s\times 2\\\\T=1.58s$

#Hence, the period of the motion is 1.58s

c. Frequency is calculated as one divided by the period of the motion.

From b above we know that the motions period is 1.58s

Therefore:

Frequency=1/period=1/1.58=0.6329Hz

#The frequency of the motion is 0.6329Hz