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Spring on a xy plane

  • Thread starter yasar1967
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A small metal block of mass m is placed on a smooth horizontal table and constrained to move along a frictionless, rectilinear groove, The block is attached to one end of a spring (of spring constant k) whose other end is fastened to a pin P.
Let length L be the equilibrium length of the spring and the perpendicular distance to the groove, The spring is now pulled a distance Xo from the equilibrium position and released. Show that, if the displacement along the groove x « t. the restoring force on
the block is proportional to x ^3 , so the motion is not simple harmonic although it will still be periodic). [Hint: For x « L, x^2 + L^2 "" L + (x^2 /2L).]




General spring and harmonic , periodic motion formulas



I thought the cosine of the F force on the spring is the responsible force i.e. restoring force and sine of F has no importance, therefore Fcosineø = -k x .. but from there I get nowhere :(
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

tiny-tim
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Hint: For x « L, √(x^2 + L^2) ~ L + x^2/2L.
Hi yasar! :smile:

What diagram have you drawn?

What is the total force in the spring at distance x along the groove?

So what is the component along the groove?

(Hint: cosine ~ x/(L + x^2/2L).)
 
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well, it's from "PHYSICS for Scientists..... by Fishbane Gasiorowicz Thornton" Tim, but I don't know how to upload a picture or a diagram :(

If you could imagine it on a xy plane, x being the groove line and L is the distance from P point to the equilibrium at y axis, I fail to see the importance of y-component of F, if any.
 
tiny-tim
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ah - I hadn't realised the question provided you with a diagram!

ok - then back to my two questions:

What is the total force in the spring at distance x along the groove?

So what is the component along the groove? :smile:
 
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>What is the total force in the spring at distance x along the groove?
Fx^2+Fy^2=F^2


>So what is the component along the groove?
Fx= Fcosineø =F x/d = F x/seqroot(x^2+L^2)

..and the above should equal to (-kx)??
 
tiny-tim
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>What is the total force in the spring at distance x along the groove?
Fx^2+Fy^2=F^2


>So what is the component along the groove?
Fx= Fcosineø =F x/d = F x/seqroot(x^2+L^2)

..and the above should equal to (-kx)??
No no no … F isn't a constant, is it? … so that doesn't get you anywhere. :frown:

The question wants you to assume that the spring has a coefficient, of µ, say, and to work out the force as a function of µ.

(When you work out proportionality later on, the µ won't matter).

So what is the total force in the spring at distance x along the groove, in terms of µ? :smile:
 
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it should be -kx cosineø (or -µx cosineø)
then
cosineø being: x/d=x/seqroot(x^2+L^2)
so
F=-(kx^2)/(seqroot(x^2+L^2))??
 
tiny-tim
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it should be -kx cosineø (or -µx cosineø)
then
cosineø being: x/d=x/seqroot(x^2+L^2)
so
F=-(kx^2)/(seqroot(x^2+L^2))??
oops … I didn't see the letter k in the question. :redface:

But shouldn't it be k times the increased length of the spring? You've used x, which is only the sideways displacement.

(and don't forget to use the hint formula in the question.)
 
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first of all thank you for the help, I appreciate.
ok, the force on the spring is F=-k(ɅL) (as ɅL is the increased length and if ɅL is zero so the force)
and that force's cosine component makes the mass M moves along the x-axis on the groove:
F=-k(ɅL)cosineø=-k(ɅL)x/seqroot(x^2+L^2)

from here where do I go???
 
tiny-tim
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Hi yasar1967! :smile:

(btw, what symbol did you intend to put before the L? It came out as a square on my iMac.)

Well, you were asked to show that the restoring force on the block is proportional to x^3.

So from now on, it's just algebra.

You have force = -k(ɅL)x/√(x^2+L^2)

So now you write ɅL in terms of L and x.

Then you use the hint formula in the question, and you get … ? :smile:
 
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that was meant to be Delta: Ʌ
:)

thanks again for the help
 
tiny-tim
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You're very welcome! :smile:

oh, and I get Delta on my iMac by pressing alt-j

Can you read that? (copy it, if you like.)

∆∆∆∆∆∆ :wink:
 

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