Simple harmonic motion on an incline

In summary, the conversation is about a physics problem involving a block on an inclined plane and a spring attached to it. The equations F = -dU/dx and F = -kx - mgsinθ are used to find the equilibrium position of the block. The solution states that the displacement of 0.075 m is from the unstretched length of the spring, not from an arbitrary position. This follows Hooke's Law which states that the force of a spring is proportional to the amount it is stretched from its unstretched length.
  • #1
PhizKid
477
1

Homework Statement


WpQ7NqB.png



Homework Equations


F = -dU/dx


The Attempt at a Solution


[itex]U = \frac{1}{2}kx^2 + mgxsin\theta \\\\
F = -(kx + mgsin\theta) \\\\
F = -kx - mgsin\theta \\\\[/itex]

We want to set the force = 0 because that's when the block is in equilibrium with no forces acting on it.

[itex]0 = -kx - mgsin\theta \\\\
x = -\frac{mgsin\theta}{k} \\\\
x = -\frac{\frac{14.0}{g}gsin\theta}{k} \\\\
x = -\frac{14.0sin(40 deg)}{120} \\\\
x = -0.075[/itex]

So since 0.075 m is the equilibrium position, it is the distance from the top of the incline to the equilibrium position, but the solution says that 0.075 m is the displacement from the position 0.450 m to equilibrium. I didn't even include 0.450 m in the equation, how can I assume that x is the distance from 0.450 to 0.075? Why is it not some arbitrary position to 0.075?
 
Physics news on Phys.org
  • #2
PhizKid said:

Homework Statement


[ img]http://i.imgur.com/WpQ7NqB.png[/PLAIN]

Homework Equations


F = -dU/dx

The Attempt at a Solution


[itex]U = \frac{1}{2}kx^2 + mgxsin\theta \\\\
F = -(kx + mgsin\theta) \\\\
F = -kx - mgsin\theta \\\\[/itex]

We want to set the force = 0 because that's when the block is in equilibrium with no forces acting on it.

[itex]0 = -kx - mgsin\theta \\\\
x = -\frac{mgsin\theta}{k} \\\\
x = -\frac{\frac{14.0}{g}gsin\theta}{k} \\\\
x = -\frac{14.0sin(40 deg)}{120} \\\\
x = -0.075[/itex]

So since 0.075 m is the equilibrium position, it is the distance from the top of the incline to the equilibrium position, but the solution says that 0.075 m is the displacement from the position 0.450 m to equilibrium. I didn't even include 0.450 m in the equation, how can I assume that x is the distance from 0.450 to 0.075? Why is it not some arbitrary position to 0.075?
The solution you refer to is correct. The 0.075 m is the amount the spring is stretched from its unstretched length.

Check Hooke's Law again.

It may say [itex]\ \ F_\text{Spring}=-k(x - x_0)\,, \ [/itex] where x is the length of the spring, and x0 is the unstretched length.

Or it may say [itex]\ \ F_\text{Spring}=-k\,x\,, \ [/itex] where x is the amount the spring is stretched (from its unstretched length).
 

Related to Simple harmonic motion on an incline

1. What is simple harmonic motion on an incline?

Simple harmonic motion on an incline is a type of motion in which an object moves back and forth along a straight line on an incline, while being pulled by a force that is proportional to its displacement from the equilibrium position.

2. What factors affect the motion of an object on an incline?

The factors that affect the motion of an object on an incline are the mass of the object, the angle of the incline, and the magnitude of the force pulling the object down the incline.

3. How does the angle of the incline affect the period of the motion?

The angle of the incline affects the period of the motion because it determines the component of the force pulling the object down the incline. A steeper incline will result in a larger component of the force, leading to a shorter period of motion.

4. What is the relationship between the period and frequency of simple harmonic motion on an incline?

The period and frequency of simple harmonic motion on an incline are inversely proportional. This means that as the period increases, the frequency decreases, and vice versa.

5. How is energy conserved in simple harmonic motion on an incline?

In simple harmonic motion on an incline, energy is conserved because the force pulling the object down the incline is a conservative force. This means that the total mechanical energy (kinetic and potential) of the object remains constant throughout the motion.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
222
  • Introductory Physics Homework Help
Replies
25
Views
321
  • Introductory Physics Homework Help
Replies
16
Views
546
  • Introductory Physics Homework Help
Replies
2
Views
684
  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
138
  • Introductory Physics Homework Help
Replies
24
Views
261
  • Introductory Physics Homework Help
Replies
2
Views
125
  • Introductory Physics Homework Help
Replies
7
Views
314
Back
Top