Doubts about the method of determining the elastic constant of a spring

[Cloruro de potasio]

Homework Statement

Some doubts about the methods of determining the elastic constant of a spring

Relevant Equations

\omega = \sqrt (k/m)

Good afternoon,I am preparing a laboratory report on the study of the oscillations of a spring and the following questions have arisen:The script asks us to represent the mass against the squared period, in this case, the slope will correspond to the spring constant divided by 4Π^2 and the ordinate at the origin with the threshold mass or effective mass of the spring. However, they also ask me to represent the angular frequency squared against the inverse of the mass, and in this case the result I obyain is strange, what does the ordinate in the origin means in this case??, what does the slope mean??

I thought the slope meant the spring constant, if we do it like that we don't take the effective mass into the account...

I will be grateful if I find this doubt.

Also, they ask me which method is more precise, the stationary or the dynamic, what do they think?

 

[KDPhysics]

Think about it like this. Let us denote the period of the spring oscillations (which I assume are linear and obey Hooke's Law with the parameters you used) by $T$ and the angular frequency by $\omega$ Then, quite clearly:
$$T^2 = \frac{4\pi^2}{\omega^2} \implies T^2 \propto \frac{1}{\omega^2}$$

Therefore, if you plot the mass against the period squared T², it will be the same as plotting the mass against the inverse angular frequency squared times a proportionality constant ($4\pi^2$).

Indeed, let's look at what the plot of your data should look like for the angular frequency plot. We know that
$\omega^2 = \frac{k}{m}$. We can denote the inverse mass by $x = \frac{1}{m}$. Plotting the inverse mass against the squared angular frequency, we get the following relation:
$$\omega^2 = kx$$
Thus, the slope will be the spring constant.
Now, let's look at what the plot of your data should look like when you plot the mass against the squared period:
$$T = 2\pi\sqrt{\frac{m}{k}}$$
Squaring both sides:
$$T^2 = 4\pi^2\frac{m}{k}$$
Thus, replacing $m = x$ since our independent variable was the mass, we get the relation:
$$T^2 = \frac{4\pi^2}{k}x$$
You can see that here, the slope of the graph will be $\frac{4\pi}{k}$. In other words, it will have the inverse slope of the angular frequency plot, times a constant $4\pi$.

 

[Cloruro de potasio]

Thank you very much, the problem is that the spring has an effective mass, which should be added to the mass that is applied on the spring to use these equations, and this effective mass is unknown.

If you develop both equations considering m = m_eff + m_applied, clearing the equation corresponding to the period we have:

m = (k/4/Π^2)*T^2 - m_0,

So representing the mass versus the squared period we obtain a slope proportional to k, and the ordinate at the origin corresponds to the effective mass. My doubt is that in this case, I do not know how to determine (taking into account the effective mass), the physical meaning of the slope and of the ordinate at the origin of the representation ω^ 2 versus 1 / m.

EDIT: Sorry, I meant ω^ 2 versus 1 / m.

 

[Cloruro de potasio]

Sorry, I meant ω^2 versus 1/m...

 

[KDPhysics]

Ok. Starting from the equation you gave in your post:
$$m = \frac{k}{4\pi^2}T^2+m_0$$
try to substitute $T = \frac{2\pi}{\omega}$

 

[Cloruro de potasio]

Thank you again,

I get the following equation:

m = k / ω^2 - m_0.

That is to say, the representation of the mass against the inverse of the frequency at the cudrado should be the result of the constant of the spring, and as ordered in origin the effective mass of this. However, if we clear ω^2 in order to obtain the equation of the line of m vs ω^2, we get ω^2 = k / (m+m_0), that is, since we don't know the value of m_0, This expersion does not make much sense, and that is why I do not know how to interpret this graphic representation.

 

[KDPhysics]

Well, to start, we can write:
$$w^2 = \frac{k}{m+m_0}$$
Then, the vertical asymptote of the graph will be equal to $- m_0$, which was equivalent to the slope in the linear relation.
You can see this by clicking here:
https://www.desmos.com/calculator
and entering the equation $\frac{1}{x+k}$, and changing the k slider.

 

[BvU]

@Cloruro de potasio ( @KCl ) can you understand how to go from @KDPhysics ' post to the intercept in your plot ?

Your plot $$m = k {T\over 2\pi}^2 - m_0 \Rightarrow\qquad {1\over\omega^2} = {m+m_0 \over k}$$ has slope $1\over k$ and intercept $m_0$. Neat !

 

[Cloruro de potasio]

Yes, I understand the representation of 1/ω^2 vs m (I think the intersection would be in m_0 / k), but the script asks me to represent ω^2 vs 1/m, and it is that graph to which I don't find such a clear physical meaning ...

By the way, in your opinion, which method is more precise, static or dynamic ...

 

[BvU]

I think the intersection would be in $m_0 / k$

You think correctly. My mistake: the y-intercept is $m_0 / k$, the x-intercept is $-m_0$.

By the way, in your opinion, which method is more precise, static or dynamic ...

Haha, by now you are the expert: you have actually done the experiment !

(did you compare $m_0$ with the mass of the spring ?)

 

[Cloruro de potasio]

Yes, from my point of view, both methods are valid as long as the measurements are made in a rigorous and precise way, perhaps, I would be more inclined to the dynamic method, since with a computer data collection system it is much more precise determine the period of the oscillations that the elongation in the static method (since in this case, no matter how much we try to leave it still, it never stops completely ...)

The effective mass of the spring turns out to be approximately 5 grams (it depends on the spring, since I did the experiment for different springs), it fits quite well to what is expected theoretically (the real mass divided by 3)