# Homework Help: Reflection of Pulses (Theoretical Problems)

1. Nov 26, 2007

### skwz

1. The problem statement, all variables and given/known data
We begin by considering the forces exerted on a ring that is connected to a spring and that is free to slide along a rod

TOP VIEW: The diagram is top view
Diagram Description (sorry don't have a pic): A pulse traveling towards a free end (has NOT reached the boundary yet), moving to the right.

Similar Pic
http://xs121.xs.to/xs121/07482/aaaasoft.gif [Broken]

Assume the ring is massless and that the rod is frictionless

i.) What is the net force on the ring? (Hint: Consider what happens to the net force on an object as its mass approaches zero)

ii) Does the net force on a massless ring depend on the acceleration of the ring? Explain

iii) What is the magnitude of the gravitational force exerted on the ring? (Recall the ring is massless.)

iv) Does the force exerted on the ring by the rod have a component that is parallel to the rod? Explain (Hint: Recall the assumptions made above)

2. Relevant equations

F = ma
F = kx (Hooke's Law)

3. The attempt at a solution

Solved?
i) Since the ring is initially at equilibrium along the rod, we assume the net force is equal to zero. Horizontal displacement is caused by the spiring force and thus the net force once the pulse reaches the boundary, the ring accelerates upward by the spring force exerted by the spring

ii) No. The acceleration of spring is entirely due to the spring force exerted on the ring by the spring. Therefore the net force is entirely due to the spring force and does not depend on the acceleration of the ring.

iii) Since the ring is at equilibrium before the pulse reaches the boundary, we can say the contact force between the rod and the ring has a equal magnitude to the gravitational force. Thus gravity has a magnitude equal to the normal force (contact force) of the rod and the ring (since ring is considered massless).

iv) No. The force exerted by the rod on the ring is a normal force that is vertical and thus orthogonal to the rod.

Next they ask for a free body diagram when the ring is farthest from the equilibrium point

For this, I assume that the vertical forces are equal in magnitude and opposite in direction. For the horizontal force, its a spring force OPPOSITE in direction of motion. Does this seem right.

Last edited by a moderator: May 3, 2017
2. Nov 26, 2007

### Astronuc

Staff Emeritus
The image cannot be seen.

Please copy it to a jpg and post it on a service like Imageshack.

As for acceleration, gravity would be a factor if the direction of motion was vertical or parallel to gravity, otherwise the acceleration will be determined by the spring force, a function of displacement and spring constant, and mass subject to the spring force.

F = kx, where k is spring constant and x is deflection from equilibrium or rest/reference position.

http://hyperphysics.phy-astr.gsu.edu/hbase/permot2.html

3. Nov 26, 2007

### skwz

Refer to last post

Last edited: Nov 27, 2007
4. Nov 26, 2007

### skwz

Refer to last post

Last edited: Nov 27, 2007
5. Nov 26, 2007

### skwz

Alright updated my answers. I'm still puzzled by iii. Seems like there is more to it.

6. Nov 27, 2007

### skwz

Refer to next post

Last edited: Nov 27, 2007
7. Nov 27, 2007

### skwz

Eureka... I think I solved this. The only problem I have is that if the spring force is the restoring force once the ring is the farthest from the equilibrium point, wont we see some sort of oscillation. Or is energy transferred to the reflected pulse, therefore the ring remains at rest?