What is the displacement of a wave on a string at a given instant of time?

Click For Summary

Homework Help Overview

The discussion revolves around determining the displacement of a wave on a string at a specific instant, based on a provided velocity graph. Participants are exploring the relationship between velocity and displacement in the context of wave motion, particularly focusing on the implications of a wave traveling at 50 cm/s.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to relate the instantaneous velocity of the string's particles to their displacement, considering integration of the velocity graph. Questions arise regarding the relevance of the wave's speed and how to account for it in calculations. Some participants express confusion about the definition of a snapshot graph and how to derive displacement from velocity.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts and questions about the relationship between velocity and displacement. Some guidance has been offered regarding the need to consider time intervals for calculating displacement, but no consensus has been reached on the exact approach to take.

Contextual Notes

Participants are grappling with the implications of the wave's speed and the concept of a snapshot graph, which may lead to varying interpretations of how to approach the problem. There is also a mention of specific points along the string and their corresponding velocities, indicating a focus on localized analysis.

ZanyCat
Messages
21
Reaction score
0
"This figure is a snapshot graph of the instantaneous velocity of the particles on a string. The wave is moving to the left at 50 cm/s.

Draw a snapshot graph of the string's displacement at this instant of time."

So it seems to be a quite basic question, where it's pretty much just an integration with working out the displacement from the velocity graph.
However I don't know how to account for the 50cm/s, and if it's even relevant at all.

I've tried considering v=λf, v=ω/k, and v=-ωA*cos(kx+ωt+θ) (plus ωt because moving to the left) but have gotten nowhere, especially because I don't know how to work out the phase constant θ, and once again don't know whether it's relevant or not.

Help would be muchly appreciated :)
 
Last edited by a moderator:
Physics news on Phys.org
50cm/s gives you a reference how long a point was in those velocity regions - you need this duration for the displacement. Integrating a velocity over a length (as a direct integration would do) does not give a length.
 
I'm stuck on this question too. Can you explain exactly how you use the 50cm/s as a reference? thanks
 
Consider x=8cm. What is its current velocity? What was its velocity before that? When did the velocity change? How far did it move in that time frame?
 
mfb said:
Consider x=8cm. What is its current velocity? What was its velocity before that? When did the velocity change? How far did it move in that time frame?

Maybe I'm confused at the definition of a snapshot graph. My understanding is that each point on the graph represents a particle's vertical velocity (since waves on a string are transverse) at that position taken at one instance of time. I'm confused on how I would calculate the displacement that corresponds to each point on the velocity graph.
 
I'm confused on how I would calculate the displacement that corresponds to each point on the velocity graph.
Did you try to follow my steps?

Here are two more intermediate step:
What is its current velocity? What was its velocity before that? How far does the wave travel in 1 millisecond? How did the velocity graph look like 1 millisecond before? When did the velocity change? How far did it move in that time frame?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
1
Views
2K
Replies
27
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K