Question

A mass suspended from a spring oscillates in simple harmonic motion at a frequency of 4 cycles per second. The distance from the highest to the lowest point of the oscillation is 100 cm. Find an equation that describes the distance of the mass from its rest position as a function of time. Assume that the mass is at its lowest point when t = 0.

Answer 1

`x=-50cm*cos(8\pi (rad)/(s)t)`

Step-by-step explanation:

The general equation for a simple harmonic motion is

`x=Acos(\omega t)`

when we take the beginning of the motion in an extreme an the oscillation. If the motion starts in the lowest point we have

`x=-Acos(\omega t)`

where A is the amplitude, w is the angular frequency, t is the time and x is the position.

In this case the amplitude is 50cm, because a complete oscillation is about 100cm.

For w we have

`\omega=2\pi f=2\pi (4s^(-1))=8\pi (rad)/(s)`

Hence, the equation is

`x=-50cm*cos(8\pi (rad)/(s)t)`

HOPE THIS HELPS!!

Answer 2

Step-by-step explanation:

Given that,

Frequency of oscillation is

f = 4cycle per second

The distance from the highest to the lowest point of the oscillation is 100 cm

this is the distance from the minimum amplitude to the maximum amplitude i.e. 2A

Amplitude of oscillation

2A = 100cm = 1m

A = 1/2

A = 0.5m

An object position in simple harmonic motion (SHM) Is modelled generally as

y = A•Sin(wt)

Or

y = A•Cos(wt)

Since the amplitude is minimum at t=0

Then, the best modelled for this is

y = A•Cos(wt)

We know that, A= 0.5m

We know that,

w = 2πf

Then, w = 2π × 4

w = 8πrad/s

Then, the position of the mass at any time becomes

y = A•Cos(wt)

y = 0.5 • Cos(8πt)