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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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
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.
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