# 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...

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.

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### 2 Answers

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

`T=2pisqrt(m/k)`

and this is related to the angular frequency by

`omega=(2pi)/T`

thus

`omega=sqrt(k/m)`

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

`E_p=1/2kx^2`

whilst the kinetic energy is

`E_k=1/2mv^2`

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

`x=sqrt((mv^2)/k)=sqrt((8xx1.5^2)/20)=sqrt(0.9)`

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

`x(t)=sqrt((mv^2)/k)sin(sqrt(k/m)t)=sqrt(0.9)sin(sqrt(2.5)t)~~0.95sin(1.58t)`

**Sources:**

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.