Simple Harmonic Motion of a 12kg mass

[Mary1910]

A 12kg mass undergoes simple harmonic motion with an amplitude of 10cm. If the force constant for the spring is 320N/m:

a) Calculate the total energy of the mass spring system.

Et=½kA^2

=½(320N/m)(0.1m)^2
=1.6 Jb) Calculate the maximum speed of the mass.

Et=½mv^2 + ½kx^2

1.6 J = ½(12kg)v^2 + 0

v^2 = (3.2/12)

= √0.2667 m/s

=0.52

=5.2 x 10^-1c) Calculate the speed of the mass when it is 5.0 cm from the equilibrium point.

And this one I'm not entirely sure...

do I use a= (kx) / (m) ?

 

[Nathanael]

c) Calculate the speed of the mass when it is 5.0 cm from the equilibrium point.

And this one I'm not entirely sure...

do I use a= (kx) / (m) ?

The acceleration (=kx/m) won't be useful.
I suggest considering the energy of the system when it is compressed by 5cm. How much is spring-potential-energy and how much is kinetic energy?

 

[Mary1910]

Im not sure how I would determine how much is kinetic energy.
Fx=-kx

=(-320N/m)(0.05m)
=-16 N

 

[Nathanael]

Fx=-kx

The units are not the same on both sides of this equation... so it can't be true.

Im not sure how I would determine how much is kinetic energy.

Well you know the total amount of energy of the system. Does that total energy depend on how much the spring is compressed?

 

[Mary1910]

Well you know the total amount of energy of the system. Does that total energy depend on how much the spring is compressed?

No, not to determine the total energy of the system, but yes the speed would be affected by that..
Would Et=½mv^2 + ½kx^2 help? If I subbed 0.05m for x? I really don't know.

Thanks

 

[Nathanael]

No, not to determine the total energy of the system, but yes the speed would be affected by that..
Would Et=½mv^2 + ½kx^2 help? If I subbed 0.05m for x? I really don't know.

Thanks

Yes that's right. The total energy doesn't change and it is given by that equation.
(In part a, the reason you used x=A to find the total energy is because that is when v=0)

 

[Mary1910]

Okay

Et=½mv^2 + ½kx^2

1.6 J=½(12kg)v^2 + ½(320N/m)(0.05m)^2

v^2=(3.2/12) + 0.04

=√0.667
v=0.82 m/s

I know this can't be correct because it is faster than the max speed of the system...could I divide the maximum speed by two since it is now 5cm from the equilibrium point? Or could I divide the total energy in half and sub that into Et=½mv^2 + ½kx^2 ?

 

[Nathanael]

could I divide the maximum speed by two since it is now 5cm from the equilibrium point?

No that wouldn't work, because the speed does not vary linearly with distance from equilibrium.

Your only mistake is that 160*0.05²=0.4 not 0.04

 

[Mary1910]

Your only mistake is that 160*0.052=0.4 not 0.04

Yeah, my mistake that was a type-o. The final answer is still the same though. But how could that answer be correct when that would mean the mass travels faster at 5cm from the equilibrium point than the calculated total max speed of the mass?

 

[Nathanael]

Yeah, my mistake that was a type-o. The final answer is still the same though. But how could that answer be correct when that would mean the mass travels faster at 5cm from the equilibrium point than the calculated total max speed of the mass?

Oh sorry that was not your only mistake. Your algebra from the second to third line is wrong.

 

[Mary1910]

Well this is pretty sad, because I don't know how I went wrong with my algebra.

Could it have been
v^2=(3.2+ 0.4/12)
=√0.3
v=0.547 m/s

but even that results in a greater value than the max speed.

 

[Nathanael]

Be more careful with your algebra. The answer will come out less than the max speed. The equation is 1.6=6v²+0.4

 

[Mary1910]

Okay I think I finally have it!

1.6=6v^2+0.4
-6v^2=-1.6+0.4
(-6/-6)v^2=(-1.2/-6)
v^2=√0.2
v=0.447
v=0.45 m/s