# Simple Harmonic Motion of a 12kg mass

• Mary1910
In summary, a 12kg mass undergoing simple harmonic motion with an amplitude of 10cm and a force constant of 320N/m has a total energy of 1.6 J. The maximum speed of the mass is 0.52 m/s, and when it is 5cm from the equilibrium point, the speed is 0.45 m/s. The total energy of the system remains constant, but the speed does not vary linearly with distance from equilibrium. Careful algebra is necessary for accurate calculations.
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) ?

Mary1910 said:
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?

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

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

Mary1910 said:
Fx=-kx
The units are not the same on both sides of this equation... so it can't be true.

Mary1910 said:
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?

Nathanael said:
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

Mary1910 said:
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)

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 ?

Mary1910 said:
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

Nathanael said:
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?

Mary1910 said:
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.

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.

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

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

## 1. What is simple harmonic motion (SHM)?

Simple harmonic motion is a type of periodic motion in which a system oscillates back and forth around a central equilibrium point, following a sinusoidal pattern. It occurs when there is a restoring force that is directly proportional to the displacement of the system from its equilibrium position.

## 2. How is the period of an object in SHM determined?

The period of an object in SHM is determined by the mass of the object and the characteristics of the spring or pendulum involved. In the case of a 12kg mass, the period can be calculated using the equation T = 2π√(m/k), where T is the period in seconds, m is the mass in kilograms, and k is the spring constant in N/m.

## 3. Can the amplitude of SHM affect the period of the object?

No, the amplitude of SHM does not affect the period of the object. The period is only dependent on the mass and the spring or pendulum. The amplitude only determines the maximum displacement of the object from its equilibrium position.

## 4. How does SHM relate to real-life systems?

SHM can be observed in many real-life systems, such as a mass-spring system, a pendulum, or even in the motion of a swinging door. Understanding SHM can help us predict and analyze the behavior of these systems and make improvements in their design.

## 5. What factors can affect the frequency of SHM?

The frequency of SHM is affected by the mass, the spring constant, and the amplitude of the oscillations. It is also affected by external factors such as friction and air resistance. Changes in these factors can alter the frequency of SHM and affect the behavior of the system.

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