# Simple harmonic motion problem

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

3. 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:

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member
2018 Award
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.

JBA
Gold Member
This small error may only be due to number of decimal places of accuracy for either V or Xmax in the calculations.

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:
haruspex
Homework Helper
Gold Member
2018 Award
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.

JBA
Gold Member
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.

haruspex
Homework Helper
Gold Member
2018 Award
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?

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?

haruspex
Homework Helper
Gold Member
2018 Award
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:
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.

haruspex
Homework Helper
Gold Member
2018 Award
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.

There is no gravitational E_p.

haruspex