- #1

ethan123

- 2

- 0

## Homework Statement

A block of mass M can slide freely (without friction) on a flat surface held at an angle [itex] \theta [/itex] to the horizontal. The block is attached to a spring of natural length l and spring constant k. The other end of the spring is fixed to a nail driven into the sloping surface.

i) At equilibrium, by how much is the spring extended?

ii) The block is set in motion directly up and down the slope (i.e. not across the slope). Show that the result is simple harmonic motion with an angular frequency which does not depend on [itex] \theta [/itex].

## Homework Equations

[tex] F=-kx [/tex]

## The Attempt at a Solution

i) To be in equilibrium the net force on the block must be zero. So using vector diagram, I got:

[tex] F_{spring}=Mgsin(\theta) [/tex]

and

[tex] F_{normal}=Mgcos(\theta) [/tex]

Then,

[tex] F_{spring}=Mgsin(\theta) =-kx [/tex]

[tex] x=\frac{-Mgsin(\theta)}{k} [/tex]ii)

I know to show simple harmonic motion need to show that:

[tex] a=-\omega^2x [/tex]

I was thinking taking the second derivative of the x found in the above part to find a, but isn't that x the distance the spring is extended when at equilibrium, not a formula for the x position?