Solving Waves on a Metal Rod with kx + Φ

  • Thread starter Thread starter LCSphysicist
  • Start date Start date
  • Tags Tags
    Metal Rod Waves
Click For Summary
SUMMARY

The discussion focuses on solving the wave equation for a metal rod represented by the formula ξ = A*cos(ωt + θ)*sin(kx + Φ). The user explores two methods to determine the boundary conditions, concluding that the stress at the ends of the rod is zero, leading to the equation cos(kx + Φ) = 0, which implies Φ = (2n-1)*π. The second method considers the node at the half-length of the rod, resulting in the equation KL/2 + Φ = n*π. The user emphasizes the need to combine both approaches for a comprehensive solution.

PREREQUISITES
  • Understanding of wave equations in physics
  • Familiarity with boundary conditions in mechanical systems
  • Knowledge of trigonometric functions and their properties
  • Basic concepts of stress and strain in materials
NEXT STEPS
  • Study the application of boundary conditions in wave mechanics
  • Learn about the properties of sine and cosine functions in wave equations
  • Research methods for solving partial differential equations in physics
  • Explore the implications of nodes and antinodes in standing wave patterns
USEFUL FOR

Students and professionals in physics, particularly those focusing on wave mechanics, material science, and mechanical engineering, will benefit from this discussion.

LCSphysicist
Messages
644
Reaction score
163
Homework Statement
I will post an image right below.
Relevant Equations
All below.
1595537783294.png

I am trying to solve this question by ξ = A*cos(ωt + θ)*sin(kx + Φ)
Anyway, the two initial terms of the product helps nothing (i think), what matters is sin(kx + Φ)
So, i tried by two ways:

First:

The stress is essentially zero on the ends, that is, something like cte*∂ξ/∂x (strain) would be zero, so
L is the length of the rod.

cos(kx + Φ) need to be 0
Φ = (2n-1)*π
kL + Φ = (2n-1)*π

It is not good.

Second:

The half length position would carry a node, so
ξ = A*cos(ωt + θ)*sin(kL/2 + Φ) = 0
KL/2 + Φ = (n)*π
and with this i can not solve without Φ in the expression.
 
Physics news on Phys.org
I assume ξ is displacement. You would find it easier with x measured from the centre of the rod.
 
  • Like
Likes   Reactions: LCSphysicist
haruspex said:
I assume ξ is displacement. You would find it easier with x measured from the centre of the rod.
Definitively is easier deal with x measured from the center of the rod, but what is wrong in the above approach?
 
LCSphysicist said:
Definitively is easier deal with x measured from the center of the rod, but what is wrong in the above approach?
It seems to me you should need to use the knowledge of the state both at the clamped point and one end, so they are not alternative methods. Rather, they should be combined into a single method.
Also
LCSphysicist said:
cos(kx + Φ) need to be 0
Φ = (2n-1)*π
I assume you meant cos(kL + Φ).
I would have thought the free end would have maximum amplitude.
On the other hand, cos((2n-1)*π)=-1, not 0, so you have effectively taken it as max.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
8K
Replies
8
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
4K
Replies
5
Views
10K