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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?

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