Tranverse velocity of a point on a string

Click For Summary

Homework Help Overview

The discussion revolves around determining the transverse velocity of a point on a string described by a sinusoidal wave equation. The original poster presents the equation governing the displacement of the wave and seeks assistance in calculating the transverse velocity at a specific time and position.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of transverse velocity and its distinction from wave speed. There is an exploration of the relationship between displacement and velocity, particularly focusing on derivatives. Some participants attempt to derive the transverse velocity using the provided wave equation.

Discussion Status

Several participants are engaged in clarifying concepts and exploring the mathematical relationships involved. There is an ongoing evaluation of the calculations presented, with some participants suggesting checks for potential errors in the evaluation process.

Contextual Notes

Participants note the importance of ensuring that units are consistent, particularly regarding radians and degrees, as this may affect the calculations. There is also a mention of the original poster's uncertainty about the initial steps in the problem-solving process.

jegues
Messages
1,085
Reaction score
3

Homework Statement



A sinusoidal wave is moving along a string. The equation governing the displacement as a function of position and time is,

y(x,t) = 0.12sin[8 \pi(t-\frac{x}{50})],

where x and y are in meters, and t is in seconds. At t = 2.4s, what is the transverse velocity of a point on the string at x = 5.0m?

Homework Equations





The Attempt at a Solution



I don't know how to get started on this one.
 
Physics news on Phys.org
jegues said:
I don't know how to get started on this one.
"Transverse" means perpendicular to the string. The "transverse velocity" is not the speed of the wave. Rather it is the velocity of a tiny point on the string itself (attached to the string).

You are given the displacement of that point on a string, using the given equation,

<br /> y(x,t) = 0.12sin[8 \pi(t-\frac{x}{50})],<br />

What's the relationship between displacement and velocity (in terms of integrals, derivatives, etc.)?
 
collinsmark said:
"Transverse" means perpendicular to the string. The "transverse velocity" is not the speed of the wave. Rather it is the velocity of a tiny point on the string itself (attached to the string).

You are given the displacement of that point on a string, using the given equation,

<br /> y(x,t) = 0.12sin[8 \pi(t-\frac{x}{50})],<br />

What's the relationship between displacement and velocity (in terms of integrals, derivatives, etc.)?

Velocity is just \frac{dx}{dt} isn't it?
 
jegues said:
Velocity is just \frac{dx}{dt} isn't it?
dx/dt is the change in position per unit time (i.e. velocity) of something along the length of the string, assuming the string lies along the x-axis.

But a point on the string itself does not move along length of the string. It moves in a perpendicular, transverse direction. Specifically, it moves in the y direction. :wink: You're looking for dy/dt.
 
collinsmark said:
dx/dt is the change in position per unit time (i.e. velocity) of something along the length of the string, assuming the string lies along the x-axis.

But a point on the string itself does not move along length of the string. It moves in a perpendicular, transverse direction. Specifically, it moves in the y direction. :wink: You're looking for dy/dt.

Okay so,

\frac{dy}{dt} = 0.12 \cdot 8\pi cos(8 \pi t - \frac{8 \pi x}{50})

When I plug the numbers in I get,

\frac{dy}{dt} = 1.6m/s

Which is still incorrect?
 
Bump, still looking for help on finishing this one off!
 
Check the evaluation, if you did not mix radians with degrees.

ehild
 
jegues said:
Bump, still looking for help on finishing this one off!
I got 2.868 m/s.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
13
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
1K