Spring sysytem: finding A and Vmax

[satutino]

There is a mass on a spring of 4 kg. The spring constant is 300 Nm. The spring is laying on a horizontal surface. (F=39.2 N)

I need to find:
1. the amplitude if the force is stretching the spring to its maximum displacement, and
2. the velocity of the object at a distance of 0.10 m from the mean position.

To find x, I used F=kx. 39.2=300x, so x=0.13 meters. Would that be the maximum displacement?

I do not know how to find the mean position of the spring. I know the spring has maximum velocity at the mean position when in motion, and I know that the spring also has maximum potential energy at the mean position when stationary.

Can someone give me a push in the right direction? Thanks!

 

[AlexChandler]

Mean position is another word for equilibrium position. This is the position the spring is at when it is not stretched or oscillating. The mean position is therefore x=0.
Yes you have found the right maximum displacement.
But actually you've got one thing wrong. You said:
"I know the spring has maximum velocity at the mean position when in motion, and I know that the spring also has maximum potential energy at the mean position when stationary."
The system actually has minimum potential energy at the mean position (it has zero potential energy
U_{spring} = \frac{1}{2} k x^2 = \frac{1}{2} k (0)^2 = 0

The velocity at a position x you can get by considering conservation of energy. Consider the initial configuration when the spring is at maximum extension. At this time we have:

K.E. = 0
U_{spring} = \frac{1}{2} k A^2

and at some later time when the spring is extended to an arbitrary position x

K.E. = \frac{1}{2} m v^2
U_{spring} = \frac{1}{2} k x^2

You can set the total initial energy equal to the total final energy, and apply that equation to the position x=.10 m

Try this and see what you get.

 

[Basher]

There is a mass on a spring of 4 kg. The spring constant is 300 Nm. The spring is laying on a horizontal surface. (F=39.2 N)

I need to find:
1. the amplitude if the force is stretching the spring to its maximum displacement, and
2. the velocity of the object at a distance of 0.10 m from the mean position.

To find x, I used F=kx. 39.2=300x, so x=0.13 meters. Would that be the maximum displacement?

I do not know how to find the mean position of the spring. I know the spring has maximum velocity at the mean position when in motion, and I know that the spring also has maximum potential energy at the mean position when stationary.

Can someone give me a push in the right direction? Thanks!

okay, firstly you have got the max displacement through Hooke's law i.e F = kx = 0.13m.
Thus you have got the Amplitude A.

Okay the total mechanical energy of the system is E = 1/2kA^2 = 2.535 Joules.

When the spring is fully pulled back the total energy is potential i.e E = 1/2kx^2 = 1/2kA^2

When the spring is past the equilibrium position all the energy is kinetic i.e E = 1/2mVmax^2 = 1/2kA^2 = 2.535 Joules.

anywhere between -A to the Equilibrium position to A will be part potential part kinetic ie E = 1/2mv^2 + 1/2kx^2.

so you know the total energy, simply substitute the displacement x, the spring constant k and the mass m into the equation to find v^2.

2.535 = 1/2mv^2 + 1/2kA^2 (substitute and solve for v).

 

[AlexChandler]

2.535 = 1/2mv^2 + 1/2kA^2 (substitute and solve for v).

This last equation should be

2.535 \normaltext{J} = \frac{1}{2} m v^2 + \frac{1}{2} k x ^2