Physics Lab: what happens if we ignore gravity?

tahayassen

So I did this lab: http://www.physics.ryerson.ca/sites/...pcs125/SHM.pdf

Essentially, we have a mass that was attached to a vertical spring with as little friction as possible. We let the mass reach equilibrium and then we applied a force on the mass. The mass oscillated with the same amplitude above and below the equilibrium position. We recorded the position of the mass and it looked like a perfect sine wave. My question is: what happens if we removed gravity? Also, we model simple harmonic motion and waves using sinusoidal functions. Can we model waves without these functions?

sophiecentaur

You could do the same experiment with the spring in space, as long as the coils of the spring were not touching (easy to do by designing the spring right or by over stretching it a bit). The period of oscillation would be the same.
The sine function comes from the solution to the equation of motion of a spring for which the restoring force is proportional to the displacement.

tahayassen

Why must the coils of the spring not touch? How come the position function is a perfect sine wave with gravity? It doesn't make sense to me. As the mass is moving up while below the y-axis, wouldn't it be slowed down by gravity? And why the mass is moving down while above the y-axis, wouldn't it be sped up by gravity?

Does the solution to the equation of motion of a spring explicitly give the sine function as the only solution? So it would be impossible to model the wave if trigonometry was not studied and the sine function was not discovered?

Doc Al

The effect of gravity, which is a constant downward force, is just to change the equilibrium position from what it would be without gravity.

sophiecentaur

If the coils touch other, the motion will be interfered with and it will no longer be SHM.
In gravity, the same force (the weight) is always in the same direction and doesn't affect the dynamic motion.

The sine function is just a way of describing a bit of geometry (at its simplest). You can relate SHM to circular motion in simple terms, as with the motion of a piston resulting from a crank going round, but, as a general principle, you can't do advanced Physics without bringing in Maths functions - the problems just can't be stated in a way that can be solved if you don't use some form of Maths. Newton, for instance, had to make up his own form of calculus before he could get anywhere.
The sine function is the solution to an equation - there isn't 'an easier one'.

tahayassen

Well, I wasn't asking for an easier one necessarily. I was just wondering if there was any alternative to the sine function.

I'm still confused to why gravity doesn't affect dynamic motion. I've uploaded a diagram.

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Doc Al

Your diagram is inaccurate. The net force at equal distances from equilibrium will be equal (in magnitude). At the equilibrium position (where the sine function equals zero) the net force will equal zero.

Realize that spring force depends on the displacement from the spring's unstretched position, not the new equilibrium position once the weight is added.

tahayassen

Is this diagram better? I think I get it now. The spring force compensates for the constant weight at ALL positions.

Thanks for the clear-up.

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Doc Al

Much better. I think you've got it. Good job.

vanhees71

The first question is answered by the theory of linear ordinary differential equations. For your case it reads
[tex]m \ddot{x}=-k x[/tex]
or
[tex]\ddot{x}=-\omega^2 x \quad \text{with} \quad \omega=\sqrt{k/m}.[/tex]
This means you look for all functions that reproduce themselves up to a factor [itex]-\omega^2<0[/itex].

There are two functions, which come immediately to mind:
[tex]x_1(t)=A \cos(\omega t), \quad x_2(t)=B \sin(\omega t).[/tex]
Since the differential equation is linear, also any superposition of the two solutions is another solution. Thus we have
[tex]x(t)=A \cos(\omega t)+B \sin(\omega t).[/tex]
One can strictly prove that this is the complete set of solutions. The two constants [itex]A[/itex] and [itex]B[/itex] must be determined from the initial conditions. Since the differential equation is of 2nd order, you must give both the initial position [itex]x_0=x(0)[/itex] at [itex]t=0[/itex] and also the initial velocity [itex]v(0)=\dot{x}(0)=v_0[/itex]. Now
[tex]v(t)=\dot{x}(t)=-A \omega \sin(\omega t) + B \omega \cos(\omega t).[/tex]
Thus we find
[tex]x(0)=A=x_0, \quad v(0)=B \omega=v_0 \; \Rightarrow B=\frac{v_0}{\omega}.[/tex]
Thus the solution, fulfilling the initial conditions, is uniquely determined to be
[tex]x(t)=x_0 \cos(\omega t)+\frac{v_0}{\omega} \sin(\omega t).[/tex]
One can also rewrite this solution into the form given on your lab sheet:
[tex]x(t)=A \sin(\omega t+\phi).[/tex]
Due to the addition theorem this is
[tex]x(t)=A [\sin(\omega t) \cos \phi+\cos(\omega t) \sin \phi].[/tex]
Thus you just have to choose [itex]A[/itex] and [itex]\phi[/itex] such that
[tex]A \cos \phi=\frac{v_0}{\omega}, \quad x_0=A \sin \phi.[/tex]
This gives
[tex]A^2 (\cos^2 \phi+\sin^2 \phi)=A^2= \left (\frac{v_0}{\omega} \right)^2 + x_0^2,[/tex]
leading to
[tex]A=\sqrt{ \left (\frac{v_0}{\omega} \right)^2 + x_0^2}.[/tex]
Further, with this value of [itex]A[/itex] the phase is given by
[tex]\phi=\mathrm{sign}(x_0) \arccos\left (\frac{v_0}{\omega x_0} \right).[/tex]

To your second question. Let [itex]x'[/itex] the coordinate of the point mass defined such that for [itex]x'=0[/itex] the spring is unstretched. Then the total force acting on the mass is the sum of the gravitational force and the force from the spring:
[tex]F=m g-k x'.[/tex]
The point, were [itex]F=0[/itex], i.e., where the force of the spring is exactly compensating the gravitational force, is given by
[tex]m g-k x_0=0 \; \Rightarrow x_0=\frac{m g}{k}.[/tex]
Thus you can write for the force
[tex]F=k(x_0-x').[/tex]
Now setting [itex]x=x'-x_0[/itex] you find the equation of motion
[tex]m \ddot{x}'=m\ddot{x}=F=k(x_0-x')=-k x,[/tex]
which is the equation of motion for a harmonic oscillator. So, as the other posters have already explained, the reason that the gravitational force drops out is due to the counting of the [itex]x[/itex] coordinate from the equilibrium point [itex]x_0[/itex].