Two Problems About Osscilations

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Homework Help Overview

The discussion revolves around two problems related to oscillations. The first problem involves determining the maximum speed of an oscillator given its mass, speed at specific displacements, and relevant formulas. The second problem concerns a block system oscillating on a spring, where the coefficient of static friction is to be found when the top block begins to slip.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore simultaneous equations based on given velocities and displacements for the first problem. There is discussion about the setup of equations and the relationship between sine and cosine functions in oscillatory motion. In the second problem, assumptions about the forces acting on the blocks and the role of friction are questioned.

Discussion Status

Some participants have provided guidance on setting up equations correctly and clarifying the relationships between variables. There is an acknowledgment of potential algebra errors and a focus on refining the approach to reach the correct answer. The second problem has seen a participant retract their previous confusion, indicating some progress in understanding.

Contextual Notes

Participants express uncertainty about their reasoning and calculations, particularly in the context of oscillatory motion. There is mention of difficulties in applying theoretical knowledge to problem-solving, highlighting the challenges faced in this area of study.

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#1- A .3kg osscilator has a speed of .954m/s when its displacement is .03m and a speed of .714m/s when its displacement is .06m. What is the Maximum speed of the osscilator? I have no idea where to start on this one all I know are the formulas I need a hint.

Formulas: [tex]v_{max} = \omega A[/tex]

[tex]x(t) = A\cos (\omega t + \phi_0)[/tex]

[tex]v_x(t) = -\omega A\sin (\omega t + \phi_0)[/tex]

#2- A small block (1kg) is on top of a bigger block(5kg) and they are osscilating on a spring of k= 50N/m with a period of 1.5s on a frictionless table. When the amplitude is increased to .4m the small block begins to slip. What is the coefficient of static friction in between the two blocks.

This didn't seem that hard to me. When the block just begins to slip, static friction would be eual and opposite to the force from the spring so:

[tex]F_{sp} = f_s[/tex]
[tex]-kx = \mu_s mg[/tex]
[tex]\frac{-ks}{mg} = \mu_s[/tex]

When I fill in the numbers here i get a [tex]\mu_s[/tex] value of 2.04 which can't be right so what's wrong with my reasoning? Also I seem to be having a lot of problems trying to teach myself osscilatory motion. I understand the theory well enough but doing the problems is another story altogether. If anyone has any advice or any good mathmatical methods to remember when doing these problems, please post them They'll be greatly appreciated.
 
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#1 - Try setting up two simulataneous equations using the velocities and displacements you are given.
 
Ok I set up my equations like this for #1:

[tex]\frac{.03}{A} = \cos\phi[/tex] and

[tex]\frac{.954}{-\omega A} = \sin\phi[/tex] If I square both sides and add these equations I get:

[tex]\frac{9X10^{-4}}{A^2} + \frac{.901}{\omega^2A^2} = 1[/tex]

since, [tex]\sin^2\phi + \cos^2\phi = 1[/tex]

Now,I do this all again with thesecond set of data and I get:

[tex]\frac{.0036}{A^2} + \frac{.510}{\omega^2A^2} = 1[/tex]

Since both of these equations equal one I can set them equal to each other. When I do this I can get the [tex]A^2[/tex]'s to cancel and i can solve for [tex]\omega[/tex]. Then I can fill in and solve for A and then [tex]v_{max}[/tex] I still get the wrong answer. Th right answer is 1.02m/s. I get .962m/s. Is my reasoning wrong or did I just make an algebra error?
 
Just Bumping the thread.
 
Im sorry, I just realizd that I posted wrong info for the second problem. Forget About that one I solved it all by myself! :biggrin: Sorry
 
I don't think you can say they each equal each other, the sum of the two equations equal one. I think what you should get is [tex]\frac{.0036}{A^2} = 1 - \frac{.510}{\omega^2A^2}[/tex] . All your other working is right so I think you should get the right answer.
 
Last edited:
Thanks a lot!
 
In #2, I presume that the spring is attached to the bottom block,
so the top block is only accelerated by friction. The spring accelerates both blocks , so the acceleration of the top block is _______ .
 

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