Question

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose it is displaced 0.125 m from its equilibrium position and released with zero initial speed. After 0.860 s, its displacement is found to be 0.125 m 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

Concepts and reason

The concept used to solve the problem is body in a Simple harmonic motion.

The equations of Simple Harmonic motion is used to determine amplitude, time period and frequency of the oscillation.

Fundamentals

The amplitude of the spring in an oscillatory motion is the maximum displacement made by the object. It is denoted by A.

Time period is the time taken by the object to complete one cycle or one oscillation. It is denoted as T.

Frequency of the object in oscillatory motion is the number of oscillations completed in one second. The frequency of the object is reciprocal of time period of the object. Frequency is expressed as follows:

Amplitude of the object is the maximum displacement. The object is displaced at 0.125m from the equilibrium position. The amplitude of the object is expressed as:

The time taken for the object from one end to the other end is 0.860 sec. The time taken to come back to the original position is given as:

Substitute 0.86 sec for and find the time period of the object.

The time period of the motion is 1.72 sec

The frequency of the oscillation is expressed as:

Substitute 1.72 sec for T and find the frequency of the oscillation.

The frequency of the oscillation is .

Ans:

The amplitude of the object is 0.125m

The time period of the motion is 1.72 sec

The frequency of oscillation is 0.581 Hz.