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[SHM] Simple Harmonic Motion

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1. Homework Statement

A disk of mass [tex]M[/tex], attached to a spring of constant [tex]k[/tex] fixed to a vertical wall can oscillate freely in a frictionless horizontal plane. A small block of mass [tex]m[/tex] is put on the disk with which there is friction under a coefficient of friction [tex]\mu[/tex]. What is the maximum amplitude of oscillation for the disk in order for the small block not to fall off?

2. Homework Equations

SHM equation for the displacement of the system:

[tex]x(t) = acos(\omega t) + bsin(\omega t)[/tex], or

[tex]x(t) = Acos(\omega t + \phi),\ A = \sqrt{a^2 + b^2}[/tex]

[tex]\omega = \sqrt{\frac{k}{(m + M)}} [/tex]

[tex]F_{friction} = \mu m g [/tex]


3. The Attempt at a Solution

So, I figured it out that there are two opposite forces acting on the small block to be taken into consideration, the friction and the elastic force (due to the oscillation of the disk).

If the elastic force is greater than the friction the block falls off the disk, but if the elastic force is equal or less than the friction the block will remain steady. And the maximum amplitude happens for the maximum elastic force which happens to be the case when it equals the friction, that is, Friction = Elastic Force.

Therefore, mathematically we can write this as

[tex]\mu m g = - k x [/tex]

[tex]\mu m g = -kAcos(\omega t + \phi) [/tex]

[tex]A = -\frac{\mu m g}{kcos(\omega t + \phi)} [/tex]

The minimum value for the denominator happens when the cosine has it's lowest value, that is, -1, therefore

[tex]A = \frac{\mu m g}{k} [/tex]

This is what I can think off, but I'm sure it is incorrect, can anyone please give me a hint?
 
Last edited:

collinsmark

Homework Helper
Gold Member
2,771
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Hello Je m'appelle,

Oh, so close!
So, I figured it out that there are two opposite forces acting on the small block to be taken into consideration, the friction and the elastic force (due to the oscillation of the disk).
There's the mistake. Draw a free body diagram. You'll find that there is only one force acting on the small block: the force of static friction. As far as the small block is concerned, it is this force, and only this force that causes it to accelerate back and forth.

You can find this by considering the maximum force possible that the block can experience without sliding. So that force is Ffriction = mμg.

Since that's the only force involved on the small block, you can calculate the small bock's maximum acceleration: ma = mμg. Solve for a

Then go back to the system as a whole. You know that the force on the spring is the combined mass (i.e. the mass of disk + small block combination) times the acceleration (and you know what the maximum a is from the above paragraph). So now take that and set it equal to the spring's force. You should be able to take it from there.
 
119
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Hello Je m'appelle,

Oh, so close!

There's the mistake. Draw a free body diagram. You'll find that there is only one force acting on the small block: the force of static friction. As far as the small block is concerned, it is this force, and only this force that causes it to accelerate back and forth.

You can find this by considering the maximum force possible that the block can experience without sliding. So that force is Ffriction = mμg.

Since that's the only force involved on the small block, you can calculate the small bock's maximum acceleration: ma = mμg. Solve for a

Then go back to the system as a whole. You know that the force on the spring is the combined mass (i.e. the mass of disk + small block combination) times the acceleration (and you know what the maximum a is from the above paragraph). So now take that and set it equal to the spring's force. You should be able to take it from there.
So,

[tex]a = \mu g [/tex]

[tex](m+M)\mu g = -k x [/tex]

[tex]A = \frac{(m+M)\mu g}{-kcos(\omega t + \phi)}[/tex]

[tex]A = \frac{(m+M)\mu g}{k}[/tex]

This?
 

collinsmark

Homework Helper
Gold Member
2,771
1,129
[tex]a = \mu g [/tex]

[tex](m+M)\mu g = -k x [/tex]

[tex]A = \frac{(m+M)\mu g}{-kcos(\omega t + \phi)}[/tex]

[tex]A = \frac{(m+M)\mu g}{k}[/tex]
Yeah, that looks like the right idea to me. :approve: (with a couple stipulations that we are evaluating things at times of maximum acceleration, maximum amplitude and such.)
 
Last edited:
119
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Yeah, that looks like the right idea to me. :approve: (with a couple stipulations that we are evaluating things at times of maximum acceleration, maximum amplitude and such.)
Thanks for the huge help collins :cool:
 

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