A frictionless spring with a 8-kg mass can be held stretched 2 meters beyond its natural length by a force of 40 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after t seconds.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

This question requires Hooke's law to solve:

`F=-kx`

where F is the force in newtons used to displace the mass by x in metres; k is the spring constant. With the force used to displace the mass, we can know the spring constant.

`k=F/x=20N/m`

Using the fact that 

F=ma

then it can be rearranged with Hooke's law to give

`a+k/mx=0`

which can also be written as 

`(d^2x)/dt^2+k/mx=0` 

The solutions to this equations take the form

`x(t)=Asin(omegat)`

and 

`x(t)=Bcos(omegat)`

where A and B are the amplitudes (maximum displacement) of the system and `omega` is the angular frequency. As x(0)=0, the first of the two equations is appropriate in this occasion.

To determine the angular frequency we need the time period of the oscillator. This is 

(The entire section contains 2 answers and 402 words.)

Unlock This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial
Approved by eNotes Editorial Team