# A question regarding a spring and perhaps even energies.

1. Sep 24, 2006

### MathematicalPhysicist

a block is acted on by a spring with a constant k and a weak force of constant magnitude f. the block is pulled distance x0 from equilbrium and released. it oscillates many times and eventually comes to rest.(the block mass is M).
1)show that the decrease of amplitude is the same for each cycle of oscillation.
2) find the number of cycles n the mass oscillates before coming to rest.

here's what i did:
w=2pi/T=sqrt(k/M)
we have kx-f=Md^2x/dt^2
we have x=Acos(wt)
where A is the amplitude.
and a=-w^2Acos(wt)
if we set t=T*n (T is the time of one cycle),n number of cycles.
then we get A(cos(2pi*n)+Mw^2cos(2pi*n))=f
from here we have: A=f/(1+Mw^2)
is this enough for the first question?

for the second question, i am kind of lost here.
i think i need to apply energies but dont know exactly how?
i mean the potential energy done by the spring is kx^2/2 and this minus the work being done by friction which is equals fx, i think that this equals the first energy potential i.e kx0^2/2, (im not sure at all it's correct) and because at the end the block is at rest kx=f, ofcourse one needs to apply t=T*n, but from here im kind of stuck, if it's even correct, is it?

for those who have kleppner's book, it's at page 196, Q4.8.

2. Sep 24, 2006

### quasar987

The question doesn't make sense to me. If the force f is constant, then the equation of motion is

$$\frac{d^2x}{dt^2}+\omega_0^2x=\frac{f}{m}$$

This has for a general solution,

$$x(t)=Acos(\omega_0t+\phi)+\frac{f}{\omega_0^2m}$$

There is no decrease in amplitude there.

Are you sure the question does not talk about a damped oscillator?

3. Sep 25, 2006

### MathematicalPhysicist

what ive typed is directly from the text, btw, how did you arrive at the last equation, for x(t)?

4. Sep 25, 2006

### quasar987

The ode to solve is a non-homogeneous one. The general solution of such an equation is the sum of the general solution to the associated homogeneous one and a particular solution to the inhomogeneous one.

The associated homogeneous equation is just the equation of the SHO, hence the first term. And it is easy to check that x=f/w²m is a solution to the inhomogeneous one, hence the second term.

5. Sep 26, 2006

### MathematicalPhysicist

so, quasar you still think the question as it's stated is not solvable?

6. Sep 26, 2006

### Sojourner01

There are at least two illogical statements in the question. Firstly, springs don't exert a constant force if they have a spring constant; the entire point of k is that it's a coefficient of the restoring force kx. Secondly, the form of the damping force isn't specified in the question, so you can't include one. Without a damping force, the spring-mass system will never stop oscillating. Maybe you're supposed to assume that the damping force is simple dynamic friction, but that's a ropey one if it's not specified.

Whoever set this question needs a kick in the arse.

7. Sep 26, 2006

### MathematicalPhysicist

here's the question the attachment, sorry for the crappy resolution, it's because of the 100 kb limitation that i scanned only with 75 dpi.
it's question 4.8.

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8. Sep 26, 2006

### quasar987

You missed two important details in retyping the question.

1° The force is of constant magnitude, which implies that it is not necessairly of constant direction!

2° The force is a friction force, which implies that the force is always acting in a direction opposite to the motion.

9. Oct 11, 2006

### MathematicalPhysicist

so is it sovlable?

10. Oct 12, 2006

### quasar987

I haven't the slightest idea. Ss far as I can tell, the x(t) solution isn't even continuous! And how could the decrease is amplitude be the same in every cycle? The force is constant, so the work done in each cycle is directly proportional to the distance covered in the cycle. And amplitude² is proportional to energy. As the amplitude decreases, the work done will be lesser and lesser, meaning a smaller and smaller decrease in energy, and thus in amplitude

This problem is a complete mystery.