# Spring on a xy plane

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 :(

## The Attempt at a Solution

tiny-tim
Homework Helper
Hint: For x « L, √(x^2 + L^2) ~ L + x^2/2L.

Hi yasar!

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).)

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
Homework Helper
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?

>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
Homework Helper
>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.

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?

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
Homework Helper
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.

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.)

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
Homework Helper
Hi yasar1967!

(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 … ?

that was meant to be Delta: Ʌ
:)

thanks again for the help

tiny-tim