A spring is attached to the ceiling

Question

A spring is attached to the ceiling and pulled 5 cm down from equilibrium and released. After 3 seconds the amplitude has decreased to 3 cm. The spring oscillates 20 times each second. Assume that the amplitude is decreasing exponentially. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

D(t) =

Answer 1

You just have to find the spring constant = k

distance D in terms of time (t) in seconds is the same as the Amplitude of the oscillating spring

from spring force equation in this situation is:

-(k)(D) = (mass under spring)(acceleration due to gravity)= (m)(9.8)

We also know the frequency 20 times it oscillates per second so the time t is just t= (1/20)= 0.05 seconds is the period of oscillation

use equation y(t) = A(cos((angular frequency)(t)+ (phase shift))

There is no phase shift since each there is only 1 wave going here

So we just need to know what the amplitude = A is and the angular frequency

A= 0.05 meters and to find angular frequency use equation angular frequency = (2(pi))/Period

and the Period is = 1/frequency = (1/20) seconds= 0.05 seconds

so the angular frequency = 125.66

We are all set: Your answer is :

y(t)= (0.05)(cos(125.66(t)))

Bam!