# Homework Help: Determining an equation for measuring frequency distribution of extended system

1. Dec 2, 2012

### Violagirl

1. The problem statement, all variables and given/known data
I have an extended system consisting of a track, a spring, a spring attached to a moving cart on one end, a string attached to the other end of the cart, and a hanging object attached to the end of the string. I need to find one equation that allows me to measure the total frequency distribution of the system. In my body diagram, I found forces of kx for the force of the spring, the tension of the string on cart in the +x direction and as a force on the hanging object in the +y direction. Finally, a force of mg on the hanging object's wieght. From doing the experiment, the frequency was found by measuring the oscillations of the spring in completing a period as a total time was taken for it to complete its cycle. I'm not sure how to find the final equation for the freqency.

2. Relevant equations

Newton second law of motion: Force = mass x acceleration

Hooke's law for the force of the spring: F(spring) = -k(constant)x X(displacement of spring)

Forces of system:

In x direction: T-kx=ma

In y direction: T-mg=ma=0, T=mg

Total forces of system: mg-kx=ma

In measuring angular frequency: ω = √k/m

For measuring frequency: f = 1/2∏√k/m

For measuring period: T = 2∏√m/k

3. The attempt at a solution

The attempts I've made so far are shown above in determining the total forces of the system. I am not sure now how to incorporate the either the angular frequency or frequency of the system as a measure of total forces into one equation. Any help is appreciated! And hope this makes sense, I'm doing all this to put together into my lab report.

2. Dec 2, 2012

### BruceW

I'm not totally certain I understand what the experiment looks like. I have attached my own drawing of what I think the experiment looks like (where the little squiggle is a spring).

From looking at your equations for forces of system, you need to specify which object you are talking about. There is the hanging object and the cart, each will have its own body diagram, so you should write out both of these, then think of a way of relating the two equations.

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3. Dec 2, 2012

### Violagirl

Your diagram shows the correct setup for the experiment, thanks for drawing that out! And when I showed my total systems of equation formula, that is how I tried to show how the cart/spring relate to the hanging object:

Total forces of system: mg-kx=ma

For this experiment, we had to determine the impact of mass on the number of osillations of the spring. For one set of trials, the mass of the hanging object was adjusted and the mass of the cart was constant. as the number of times that the spring attached to the cart would oscilliate and timing how long it took for it complete one cycle. For a second set of trials, the mass of the hanging object was kept constant as the mass of the cart was adjust and the the number of times that the spring would oscillate was recorded again and in timing it again for how it took to complete a cycle.

I'm not sure from here how to come up with a single equation that shows how to find a predicted frequency to the forces equation shown above. In my previous post, I showed what each formula in simple harmonics to a spring correlates to in finding angular frequency, frequency, and period but I'm not sure how to set it up to one equation.

4. Dec 2, 2012

### Violagirl

Otherwise, in doing some research online on this, since the total forces of the systems in the system equal:

Total forces of system: mg-kx=ma

Then the differential equation can be found to show that:

Total forces of system: mg-kx=ma=m d2x/dt2=-kx

In showing second order differential equation as a function of time, it can be shown as:

x(t)=A cos(ωt+δ)

A=amplitude, ω=angular frequency, t=time, δ=angle

In the situation for this experiment, it hadn't looked like an angle had been measured.

Otherwise, I'm not sure how else to go about relating Newton's Second Law of Motion to measuring the oscillations of the spring in this system.

5. Dec 2, 2012

### BruceW

This equation is almost right, but the m on the left hand side and the m on the right hand side should different things. When you were calculating the forces, you wrote T=mg, but this is not correct, because the hanging mass will also undergo oscillations.

Remember that you have 1 equation for the vertical motion of the hanging block and 1 equation for the horizontal motion of the cart. So you need to use different values for m in each equation.

6. Dec 2, 2012

### Violagirl

Hmm, ok, so going back to looking at the equation in the x direction, we have T-kx=ma

And in the y direction, T-mg=may=0.

Before, I was thinking that T could become mg so it could be substituted into the x equation to make it into one eqaution.

So then, (although this does not feel right), would mass have to be cancelled on either side of my original equation then?

So then we'd have, g-kx=a. But then we have no way to measure differences in mass....

I'm confused.

Otherwise, I know that me need the mass to calculate the angular frequency, ω=√k/m. I'm not sure how to go from one step to the next.

7. Dec 2, 2012

### Violagirl

Or if we're trying to come up with an equation that relates both masses of the hanging object and cart to the frequency of the spring, would we not have to add both the mass of the hanging object (m1) and cart (m2) together? I'm not sure...

8. Dec 3, 2012

### haruspex

No, T won't be equal to Mhg (Mh being mass of hanging object), since that object accelerates.
You can either do the standard thing of treating each mass separately, connected by equal and opposite T and a:
T-kx=Mcax
T-Mhg=Mhay
ay = -ax
or finesse it by taking the sum of the masses as the inertial mass but Mh as the gravitational mass.

9. Dec 3, 2012

### Violagirl

In going back to look it, I found that it would make sense to take the sum of the masses in having an impact on the frequency of the spring. So therefore, I found that:

f=2ττ√mcart+massweight/k

I'm not sure what to do with the +y direction though in correlation to the +x or to Newton's second law...

10. Dec 3, 2012

### BruceW

The nice thing about this problem is that if you didn't realise that the effective inertial mass is equal to the sum of the masses, then you could use Newton's laws to solve this problem, and you would find out that the effective inertial mass is equal to the sum of the masses.

Since you have already realised that the effective inertial mass is equal to the sum of the masses, you don't need to do any of Newton's laws. (unless your teacher is expecting you to show why the effective inertial mass is equal to the sum of the masses).

p.s. I think you missed out a 1/ in your equation for f