A question regarding a spring and perhaps even energies.

Maybe you should try to solve it step by step by using the book's instructions, since they seem to make sense to you. Or ask the teacher for help. Sorry I can't help more.In summary, the conversation is discussing a block acted on by a spring with a constant k and a weak force of constant magnitude f. The block is pulled distance x0 from equilibrium and released, causing it to oscillate many times before eventually coming to rest. The first question asks to show that the decrease in amplitude is the same for each cycle of oscillation, while the second question asks to find the number of cycles the mass oscillates before coming to rest. However, there are discrepancies and missing information in the question, making it difficult to
  • #1
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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 don't 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 I am kind of stuck, if it's even correct, is it?

for those who have kleppner's book, it's at page 196, Q4.8.
btw, the answer is n=1/4[kx0/f-1]
 
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  • #2
The question doesn't make sense to me. If the force f is constant, then the equation of motion is

[tex]\frac{d^2x}{dt^2}+\omega_0^2x=\frac{f}{m}[/tex]

This has for a general solution,

[tex]x(t)=Acos(\omega_0t+\phi)+\frac{f}{\omega_0^2m}[/tex]

There is no decrease in amplitude there.

Are you sure the question does not talk about a damped oscillator?
 
  • #3
what I've typed is directly from the text, btw, how did you arrive at the last equation, for x(t)?
 
  • #4
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
so, quasar you still think the question as it's stated is not solvable?
 
  • #6
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
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
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
so is it sovlable?
 
  • #10
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.
 

Related to A question regarding a spring and perhaps even energies.

1. How does a spring store and release energy?

A spring stores and releases energy through its ability to deform and return to its original shape. When a force is applied to a spring, it stretches or compresses, storing the energy in the form of potential energy. This potential energy is then released when the force is removed and the spring returns to its original shape.

2. What factors affect the amount of energy stored in a spring?

The amount of energy stored in a spring is affected by the spring's stiffness, or spring constant, and the distance it is stretched or compressed. The greater the stiffness and the greater the distance, the more energy will be stored in the spring.

3. Can a spring have potential and kinetic energy simultaneously?

Yes, a spring can have both potential and kinetic energy at the same time. When a spring is being stretched or compressed, it has potential energy. Once released, the potential energy is converted into kinetic energy as the spring moves back to its original shape.

4. How does the mass of an object affect the energy stored in a spring?

The mass of an object attached to a spring does not affect the amount of energy stored in the spring. However, the mass of the object can affect the distance the spring is stretched or compressed, which in turn can affect the amount of energy stored in the spring.

5. Can a spring store an infinite amount of energy?

No, a spring cannot store an infinite amount of energy. There is a maximum amount of energy that can be stored in a spring, which is determined by the spring's stiffness and the distance it is stretched or compressed. Once this maximum is reached, the spring will not be able to store any more energy.

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