Simple harmonic motion problem

In summary, the system has a kinetic energy of 128 x 10^-4 at the equilibrium point and a potential energy of 1/2 * k * Xmax^2.
  • #1
NooDota
68
0
I translated it, so the notation may not be so great.

You have a vertical spring with a spring constant of 16 N.m^-1, we attach to its lower end a body with a mass of 0,4 KG and form a system which oscillates on a straight line with length of 8cm, we assume that at the start of the time, the body is in its equilibrium point and moving in the negative direction (the axis is directed downwards).

1. Find the period for oscillations and the displacement function.

2. Find the velocity and momentum function, calculate them at t = 1/2s. Calculate the kinetic energy at the same time. 2. Homework Equations

The Attempt at a Solution

I didn't have a problem with the first question, I got T=1s and the function for displacement: x=0.04 cos(2*π*t+π/2) What I'm having problem with is the kinetic energy,

if I calculate it by:

1/2 * m * v^2 I get 125*10^-4

Because v at t = 1/2 is 25*10^-2.

If I calculate it by:

1/2 * k * Xmax^2 I get 128*10^-4


Because at t=1/2, the mechanical energy is all kinetic energy.What am I doing wrong? Which answer is correct, why is there a small 3 unit difference?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
NooDota said:
If I calculate it by:

1/2 * k * Xmax^2 I get 128*10^-4
Not sure how you are calculating the KE via the spring's PE, but I suspect you are forgetting a difference in gravitational PE between two situations.
 
  • #3
This small error may only be due to number of decimal places of accuracy for either V or Xmax in the calculations.
 
  • Like
Likes NooDota
  • #4
Nevermind, I forgot that I used a π approximation when calculating V, hence the small difference. Sorry.

I guess I should stick to 1/2*k*Xmax^2 because it's more accurate.Also, haruspex, as I said in the original post, the entire energy is only kinetic since the body is at the equilibrium point, that's why I used 1/2*k*Xmax^2
 
Last edited:
  • #5
NooDota said:
Nevermind, I forgot that I used a π approximation when calculating V, hence the small difference. Sorry.

I guess I should stick to 1/2*k*Xmax^2 because it's more accurate.Also, haruspex, as I said in the original post, the entire energy is only kinetic since the body is at the equilibrium point, that's why I used 1/2*k*Xmax^2
1/2kxmax2 is the elastic potential energy when the body is at its lowest point, yes? In getting from there to the equilibrium point work is done against gravity. That will be about equal to your discrepancy.
 
  • #6
For Example: For a .4 Kg Mass / 9.8 = .0408163265 as limited by my calculator's 10 digit limit on accuracy; and using that value rather than .04 for the mass in your first KE calculation results in KE = 127.551 X 10^-4 rather than your 125 x 10^-4 answer.
 
  • #7
JBA said:
For Example: For a .4 Kg Mass / 9.8 = .0408163265 as limited by my calculator's 10 digit limit on accuracy; and using that value rather than .04 for the mass in your first KE calculation results in KE = 127.551 X 10^-4 rather than your 125 x 10^-4 answer.
Why would you divide the mass by g?
 
  • #8
haruspex said:
1/2kxmax2 is the elastic potential energy when the body is at its lowest point, yes? In getting from there to the equilibrium point work is done against gravity. That will be about equal to your discrepancy.

Still no idea what you're talking about.

The discrepancy is because I used 4π=12.5

Also no idea why g is related here?
 
  • #9
NooDota said:
Still no idea what you're talking about.

The discrepancy is because I used 4π=12.5

Also no idea why g is related here?
Ok, I see the confusion. Your xmax is displacement from the equilibrium point, not the extension of the spring. As I said in post #2, I was not sure how you were using a formula based on kx2 to get the KE. It looked like you were using spring PE, but it seems you weren't.

[edit: I still don't know on what basis you used that equation. It isn't the extra spring PE corresponding to the extra extension either.]

Of, course, the period is not exactly 1s, it's ##\frac{\pi}{\sqrt{10}}##. Correspondingly, it is not so much that you approximated 4pi as 12.5, but rather that you effectively approximated 4sqrt(10) as 12.5.
 
Last edited:
  • #10
haruspex said:
Ok, I see the confusion. Your xmax is displacement from the equilibrium point, not the extension of the spring. As I said in post #2, I was not sure how you were using a formula based on kx2 to get the KE. It looked like you were using spring PE, but it seems you weren't.
E = E_k + E_p

At x = 0, E_p = 0

Thus E = E_k

1/2 * k * Xmax^2 = E_k

I don't really see what's wrong in this.
 
  • #11
NooDota said:
E = E_k + E_p

At x = 0, E_p = 0

Thus E = E_k

1/2 * k * Xmax^2 = E_k

I don't really see what's wrong in this.
The complete energy expression would include a gravitational PE term. That will have a different value at max KE from its value at max elastic PE. On the other hand, the value, 4cm, you substituted for xmax to find max Ek is not the maximum extension. It is the difference between the maximum extension and the equilibrium extension. As it happens, these two errors(?) cancel.
If z is the actual extension and z0 the equilibrium extension then the delta in elastic PE is 1/2 k (z2-z02). If x=z-z0 then this becomes 1/2 k (x2+2xz0). But kz0=mg, so it becomes 1/2 kx2+mgx. The mgx cancels the gravitational PE term.

Now, maybe you understood all this, and it's how you arrived at the equation, but that's not evident so far.
 
  • #12
There is no gravitational E_p.
 
  • #13
NooDota said:
There is no gravitational E_p.
It's a vertical spring with a mass bobbing up and down. How can there not be gravitational PE?
 
  • #14
It just ends up being the same numbers as before? It doesn't make a difference if you add it or not?
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object moves back and forth in a regular pattern. This motion is caused by a restoring force that is proportional to the displacement of the object from its equilibrium position.

2. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A cos(ωt + φ), where x is the displacement of the object from its equilibrium position, A is the amplitude, ω is the angular frequency, and φ is the phase constant.

3. What is the difference between simple harmonic motion and oscillatory motion?

Simple harmonic motion is a specific type of oscillatory motion, in which the restoring force is proportional to the displacement from equilibrium. Oscillatory motion, on the other hand, refers to any motion that repeats itself in a regular pattern.

4. How do you calculate the period of a simple harmonic motion?

The period of a simple harmonic motion is calculated using the equation T = 2π/ω, where T is the period and ω is the angular frequency. The period represents the time it takes for one complete cycle of the motion to occur.

5. What are some real-life examples of simple harmonic motion?

Some real-life examples of simple harmonic motion include the swinging of a pendulum, the motion of a mass attached to a spring, and the vibrations of a guitar string. Many other systems can also exhibit simple harmonic motion, such as the motion of a mass on a frictionless surface or the motion of a mass attached to a rotating wheel.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
395
  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
855
Replies
13
Views
309
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
834
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
844
Back
Top