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

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)

#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

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .