# Homework Help: Simple Harmonic Motion and spring displacement

1. May 2, 2012

### NeuronalMan

1. The problem statement, all variables and given/known data
An object of mass m is suspended from a vertical spring of force constant 1800 N/m. When the object is pulled down 2.50 cm from equilibrium and released from rest, the object oscillates at 5.500 Hz. a) Find m, b) Find the amount the spring is stretched from its unstressed length when the object is in equilibrium, c) Write expressions for the displacement x, the velocity v and the acceleration a as functions of time t.

2. Relevant equations
For simple harmonic motion, we have that

x=A*cos(ωt + δ), where δ is the phase constant.

3. The attempt at a solution
For this particular problem, I am only concerned with c), and moreover finding the phase constant δ. Actually, not finding it, but to understand why it should exist in the first place, i.e., a non-zero phase constant (we can find it by solving the equation set x=A*cos(ωt + δ) and v=-ωA*sin(ωt + δ) for δ).

I gather that if a particle or an object is released from rest, that x=A, hence δ=0, which to me makes the most sense by looking at an x/t graph, seeing that it's at its greatest value at t=0, so there can be no phase shift. My intuition tells me that the same thing should apply to an object hanging from a spring. But since apparently it doesn't, I am led to think that gravity can cause the object to move past the initial point x, to which the spring is elongated, as there is seemingly no other explanation for this. As a matter of fact, this is the only reason I can think of, and it's the only thing that's different from a situation in which a spring is being pulled horizontally. Yet still, this difference isn't mentioned in the book, so I would appreciate someone explain this to me.

Thank you.

Last edited: May 2, 2012
2. May 2, 2012

### dikmikkel

You could do a Euler-Lagrange approach with energies(simple way if you know the metod)? or you could write down newtons laws of motion(just as simple) and solve the corresponding differential equation for v(integrate once?) and x integrate twice.
But don't just use a formula which is valid under other circumstances than for your problem.

3. May 2, 2012

### dikmikkel

In fact this is a sort of driven oscillation by gravity with the solutions:
$y(t) = c_1\sin(\sqrt{\dfrac{k}{m}}t)+c_2\cos(\sqrt{\dfrac{k}{m}}t) + mg/k$

4. May 2, 2012

### NeuronalMan

dikmik, thank you for your response. I just had another peek in the solutions manual, and interestingly, they've used A=2.50cm=x0 for the amplitude later in the solution. But then, x=A×cos(ωt + δ), solved for δ, should be cos(δ)=1 → δ=0, which is self-evident, and not δ=arctan(-v0/ωx0)=arctan(0), v0=0 → δ=pi, as stated in the solutions manual. I don't get this. I don't understand this.

p.s. This problem is from Physics for Scientists and Enginners (Tipler et al.) if anyone is familiar with the textbook.

5. May 2, 2012

### NeuronalMan

I just realized they've used x=-2.50 cm, but as far as I am concerned the solution to the equation x=A×cos(ωt + δ) would still be 0. Although, if we treat the amplitude as an absolute value, we get pi, the same answer as yielded by the other solution. So it was merely a question of the sign. Actually, in the solution they've gotten rid of the pi and simultaneously made the expression negative, which is perfectly legitimate. So my hunch that the amplitude would be affected by gravity was a dead end according to the solutions manual. I still think it would affect the motion of the object, as suggested by dikmikkel it's a driven oscillation, that's seemingly 'negligible' according to the solutions manual.

6. May 2, 2012

### dikmikkel

Hmm but you are ignoring the gravity it seems to me, their surface is perpendicular to the earth i would guess, but now you have to use newtons laws to solve the problem and som knowledge about differential equations. And negligble? maybe but i meant the gravity drives the oscillations at first only to reach equilibrium later.
The point is: Analyse your problem with a well defined coordinate system(where is origo?) and write down the dynamical equations(newtons laws) and solve them for Y for example. Then you can find V by realising : a = dv/dt

7. May 2, 2012

### Staff: Mentor

Perhaps your intuition is failing you here? When the mass is at rest at the equilibrium point the net force acting on that mass (from the spring and gravity) must be zero. That's essentially the definition of "equilibrium point".

When the mass is stretched past the equilibrium point and held at rest there is net force due to the spring and gravity that is being countered by whatever is holding the mass at rest in that stretched position. If the mass is then released it must move in the direction of the restoring force -- it cannot move further from the equilibrium point if its kinetic energy is zero and the net force is accelerating it towards equilibrium.

When the mass is allowed to oscillate freely, whenever it is away from the equilibrium point it is being accelerated towards that equilibrium point. This implies that the kinetic energy of the mass is being 'traded', via the acting force, with whatever is providing that force. In this case it is the spring force and gravity, two conservative forces. So energy is being traded between kinetic and potential forms.

One way to confirm that the initial stretched position does in fact represent the greatest amplitude of the oscillation is to use an energy conservation approach. Identify where energy can reside in the system and write the energy conservation equation. Note that potential energies are relative to some reference point so a thoughtful choice of that reference point will make life easier mathematically speaking.

Last edited: May 2, 2012
8. May 2, 2012

### NeuronalMan

gneill, thank you for your response. Not sounding pejorative I should point out a few things. This is sort of exactly what I was getting at, so I think you might have misunderstood what I wrote. You're right I didn't consider the energies, but intuitively, I thought the amplitude would equal x at the point to which it was stretched. Initially, I tried solving this by using the equation for simple harmonic motion, as aforementioned. The reason why my solution contained a phase δ=0, was that I treated the distance as positive, whereas if we use x=-2.50 cm and A=abs(-2.50cm), then we get pi, which basically means that we're dragging the object 'below' the x-axis.

However, the root of my confusion was that in the solutions manual, they solve this problem by solving the equation set x=A*cos(ωt + δ) and v=-ωA*sin(ωt + δ), as to eliminate A, while in my opinion, it's easier to solve by substituting for x0 and A in the original equation x=A*cos(ωt + δ). Now, since they didn't do that, it led me to think that A was unknown, i.e., different from x0, which again made me wonder if the object could actually go past x0, which is, on the other hand, quite counter-intuitive.