Block attached to spring on slope

In summary, the conversation discusses a problem involving a block sliding on a flat surface at an angle with a spring attached to it. The first part of the conversation focuses on finding the distance the spring is extended when the block is at equilibrium. The second part discusses showing simple harmonic motion and finding the formula for acceleration as a function of x. The final part clarifies that the x in the formula represents the distance from equilibrium, not the overall extension of the spring.
  • #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?
 
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  • #2
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.
 
  • #3
Thanks for the reply

So letting [itex] x_0 [/itex] be the distance from equilibrium.
[tex] F=kx [/tex]
[tex] ma=kx [/tex]
[tex] a=\frac{k}{m}x [/tex]
And since [itex] \omega =\sqrt{\frac{k}{m}}[/itex]
[tex] a=\omega^2x [/tex]
And the x found in part i) would be the distance from the natural length of the spring to equilibrium.
Does this make sense?
 
  • #4
No - if you just let the box rest at equilibrium, what is it's acceleration?
 
  • #5


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}
 

Related to Block attached to spring on slope

What is a block attached to spring on slope?

A block attached to a spring on slope is a common physics experiment that involves placing a block on a slope and attaching a spring to the block. The block is then pulled down the slope by the force of gravity and the spring, which is stretched, exerts a restoring force on the block.

What is the purpose of this experiment?

The purpose of this experiment is to study the relationship between the force of gravity, the angle of the slope, and the spring constant of the attached spring. It also helps to understand the concept of Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement.

What factors affect the motion of the block on slope?

The motion of the block on the slope is affected by the angle of the slope, the mass of the block, the spring constant, and the force of gravity. Additionally, factors such as friction and air resistance can also play a role in the motion of the block.

How can the spring constant be calculated in this experiment?

The spring constant can be calculated by measuring the displacement of the spring from its equilibrium position and the force exerted by the spring. The spring constant is equal to the ratio of these two values.

What are some real-world applications of this experiment?

This experiment has various real-world applications, such as in designing shock absorbers for vehicles, understanding the behavior of elastic materials, and studying the motion of objects on inclined planes. It also has applications in fields like engineering, architecture, and sports science.

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