Spring & SHM

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  • #1
kel
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Hi,

I've got a problem to solve, but I'm really new to this topic and I'm having a bit of trouble knowing where to start.

Here's the question I'm working on:

A mass = 1.5 kg attached to a spring with k=8 N/m on a horizontal surface with friction force -b(dx/dt) where b = -230g/s

The mass is pulled a distance 12cm and then released.

1- What is the time required for the amplitude of the oscillations to drop to 4cm?
2- How many oscillations are made by the mass in this time?

Could anyone point me in the right direction?

Thanks
 

Answers and Replies

  • #3
kel
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Hi

That looks useful, but I have a few of the formulas already, I'm just not sure how to go about tackling the problem.

I have the Fundamentals of Physics book by Halliday,Resnick,Walker, but it doesn't really make much sense and I can't find any examples with friction.
 
  • #4
Hootenanny
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Begin by looking at forces.
 
  • #5
kel
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Still not sure. I don't want the answers to this problem, I'm just not sure what I'm doing.

I was doing something along the lines of:

x(t) = 0.12 cos (2.3)t

I need to know how long it takes to get to 4cm, so

t = 0.04/0.12 cos (2.3) = 0.33 seconds

Is this right because it doesn't look it to me? How do I take the friction into account?
 
  • #6
Hootenanny
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Try approaching it from a conservation of energy viewpoint. I haven't had chance to work through this problem yet (coursework pressure :cry: ), but I will try and have a look at it today. :smile:
 
  • #7
kel
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ok, I've found an example, how does this look:

T= 0.96 s

after messing around with a formula we get:

-bt/2m = Ln 1/3 (1/3 because 4 s is 1/3 of 12)

therefore, t = 14.3 seconds

and the number of oscillations is:

t/T = 14.3/0.96 = 14.9 oscillations
 
  • #8
Hootenanny
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Did you use the 'underdamping' formula? Can you just verify that 230g/s means 230 grams per second not 230 * 9.81 per second.
 
  • #9
kel
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it's 230 grams per second.

Think part b is wrong tho - some of my colleagues have 5.27 oscillations or there abouts.
 
  • #10
Hootenanny
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Can you show your working for the time period?
 
  • #11
Hootenanny
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I've just worked through the problem now. I get the same answer as you for the time taken for the amplitude to fall to 0.04m. However, I get 5.27 oscillations, as your colleagues did. I'm not sure what you've done to find the time period because you haven't shown your working. Try calculating the time period again.

Remember: The time period of any mass spring system is given by;
[tex] T = 2\pi\sqrt{\frac{m}{k}}[/tex]

Please post your working :smile:
 
  • #12
kel
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I think I went wrong with it because I found a formula which said t/T

which would give : 14.3/0.96 = 14.9

I guess I just used the wrong formula.

Thanks for your help
 
  • #13
Hootenanny
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T is time period, t/T is correct, you just calculated the wrong time period. Use the formala I gave about to calculate time period. :smile:
 
  • #14
kel
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You know what, brackets on calculators can be sooooo annoying !

That's why my answer was wrong, it calculated the root of 1.5 and divided that by 8 instead of treating 8 as part of the root.

Great, I've got it now.

Thanks again
 
  • #15
Hootenanny
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No problem :smile:
 
  • #16
kel
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Would you mind looking at another question that I have??

The amplitude of a lightly damped oscillator decrease by 3% each cycle of oscillation. What fraction of the total energy of the oscillator is lost during each cycle?

Now I now that the total energy = Ke + Pe = 1/2mv sqrd + 1/2kx sqrd

(actually is there an equation editor on this forum, cos it's not easy writing equations on here?)

Anyway, i have the equs for the energy, but am not sure about the 3% part. There are no values, so I guess the answer will be in equation form.

How do adapt the equation to take into account 3% of the amplitude? Do I just multiply the x variable by 3/100 ?
 
  • #17
Hootenanny
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kel said:
Would you mind looking at another question that I have??
Of course not!

kel said:
(actually is there an equation editor on this forum, cos it's not easy writing equations on here?)
Yes, it's a mark-up language called [itex]LaTeX[/itex] :smile: there's a tutorial on it here: https://www.physicsforums.com/showthread.php?t=8997

If you take the energy when the system is at amplitude, you know that the system is at rest, i.e. KE=0, therefore, you can say that for one cycle;

[tex]\frac{1}{2}k(1)^2 = \frac{1}{2}k \left( \frac{97}{100} \right) ^2 + E_{lost}[/tex]
 
  • #18
kel
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Ah, so you'd work it out for 97% instead of 3%.

Makes sense!
 
  • #19
Hootenanny
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That's the way I'd do, that way the energy lost is just the bit that's left over! :smile:
 
  • #20
kel
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Thank you !! :smile:
 

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