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:


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.


Using the fact that 


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


which can also be written as 


The solutions to this equations take the form




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 


and this is related to the angular frequency by




At the maximum displacement, the potential energy is at its highest, and the kinetic energy is zero. The potential energy of a harmonic oscillator is


whilst the kinetic energy is 


By equating these two, we have that the maximum displacement of the oscillator is


Thus this gives the final statement of the displacement of the mass as


Approved by eNotes Editorial Team
An illustration of the letter 'A' in a speech bubbles

m = 8 kg.

The first data taht it can be stretched to 2 meters by a force of 40 newtons is given to find the value of k.

F = kx

`40 = k * 2`

k = 20.

Let's apply newton's second law to the movement of the spring.

`F = ma`

`-kx = m(d^2x)/(dt^2)`

`(d^2x)/(dt^2) = -(20/8)*x`

`(d^2x)/(dt^2) = -2.5x`

`(d^2x)/(dt^2) = -omega^2*x`

where, `omega = sqrt(2.5)`

We know the intial conditions at t =0,

`x(0) = 0`

`(dx)/(dt) (0)= +1.5` (this must be positive, otehrwise there cant be harmonic motion)

we know for harmonic motion, `x(t) = Acos(omegat-phi)`

`(dx)/(dt) = -Aomegasin(omegat-phi)`

`(d^2x)/(dt^2) = -Aomega^2cos(omegat-phi)`

this gives you, ` ` `(d^2x)/(dt^2) = -omega^2x`

by applying initial conditions, x(0) = 0

`0 = Acos(omega*0-phi)`

`cos(phi) = 0`

`phi = pi/2`

`(dx)/(dt) = 0`

`1.5 = -Aomegasin(omega*0-pi/2)`

`1.5 = Asqrt(2.5)sin(pi/2)`

`A= 1.5/sqrt(2.5) = 3/sqrt(10)`

Therefore x is,

`x(t) = 3/sqrt(10)cos(omegat-pi/2)`

This gives you the location of mass at time t.



See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team