Block attached to spring on slope

[ethan123]

Homework Statement

A block of mass M can slide freely (without friction) on a flat surface held at an angle $\theta$ 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 $\theta$.

Homework Equations

$$F=-kx$$

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:
$$F_{spring}=Mgsin(\theta)$$
and
$$F_{normal}=Mgcos(\theta)$$

Then,

$$F_{spring}=Mgsin(\theta) =-kx$$
$$x=\frac{-Mgsin(\theta)}{k}$$ii)
I know to show simple harmonic motion need to show that:
$$a=-\omega^2x$$
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?

 

[Simon Bridge]

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?

...You mean you don't know what the "x" in the formula you have to demonstrate means?
You should ask then.

But does it matter?
Your process is the same: derive the formula for acceleration as a function of x and see if it matches. If you like you can keep track of both the overall extension and the extension at equilibrium and see which one matches the formula you have to show.

 

[ethan123]

Thanks for the reply

So letting $x_0$ be the distance from equilibrium.
$$F=kx$$
$$ma=kx$$
$$a=\frac{k}{m}x$$
And since $\omega =\sqrt{\frac{k}{m}}$
$$a=\omega^2x$$
And the x found in part i) would be the distance from the natural length of the spring to equilibrium.
Does this make sense?

 

[Simon Bridge]

No - if you just let the box rest at equilibrium, what is it's acceleration?

 

[HalfLostAndSearching]

I would like to clarify a few things before providing a response to the given content. Firstly, I would like to know the context of this problem and the purpose of solving it. Is it a theoretical exercise or a practical experiment? This information will help me provide a more accurate and relevant response.

Assuming this is a theoretical exercise, here is my response:

i) To find the extension of the spring at equilibrium, we can use the formula for Hooke's Law, F=-kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. In this case, the equilibrium position is when the block is at rest on the slope. Therefore, the net force acting on the block must be zero. As you have correctly mentioned, the forces acting on the block are the weight (Mg) and the spring force (kx). At equilibrium, these forces must be equal and opposite, thus we get:

Mg sin(\theta) = -kx

Solving for x, we get:

x = -Mg sin(\theta)/k

Therefore, the spring will be extended by a distance of Mg sin(\theta)/k at equilibrium.

ii) To show that the motion of the block is simple harmonic, we need to show that the acceleration of the block is directly proportional to its displacement from the equilibrium position, and the acceleration is always directed towards the equilibrium position. In this case, the block is moving up and down the slope, so we need to consider the component of the gravitational force along the slope, which is Mg sin(\theta). This force is balanced by the spring force, -kx. Thus, we can write the equation of motion as:

Mg sin(\theta) = -kx

Taking the second derivative with respect to time, we get:

Mg cos(\theta) \dot{\theta}^2 = -k \ddot{x}

As we can see, the acceleration is directly proportional to the displacement (x) and is always directed towards the equilibrium position. This shows that the motion is simple harmonic. Now, to find the angular frequency, we can rearrange the above equation as:

\ddot{x} = -\frac{Mg}{k} cos(\theta) \dot{\theta}^2

Comparing this with the standard equation for simple harmonic motion, \ddot{x}