Calculating Speed and Max Transverse Velocity of a Transverse Wave

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Homework Help Overview

The discussion revolves around a homework problem involving the analysis of a transverse wave described by the equation y = 2.28sin(0.0276pi x + 2.42pi t). Participants are tasked with finding various properties of the wave, including amplitude, wavelength, frequency, speed, and maximum transverse speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of speed using the relationship w/k and question the determination of maximum transverse speed. There is exploration of the relationship between displacement and velocity functions, particularly the role of the cosine function in finding maximum values.

Discussion Status

Some participants express confidence in their calculations, while others seek clarification on the nature of the velocity function and its maximum values. There is an ongoing exploration of how to derive the maximum transverse speed from the wave equation.

Contextual Notes

Participants are navigating through the implications of differentiating the wave function and the assumptions regarding the nature of velocity as a function of position and time. There is a noted uncertainty about the correct interpretation of maximum speed in relation to the wave's properties.

croyboy
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Please help!

Having a little difficulty with a homework question and concept

Homework Statement


y = 2.28sin(0.0276pi x + 2.42pi t)

Find the amplitude, wavelength, frequency, speed and maximum transverse speed.



The Attempt at a Solution



Definitely correct answers

A = 2.28cm
Wavelength = 72.46 cm
Frequency = 1.21 Hz

Questionable answers

Speed = 87.68 cm/s
Max transverse speed = ?


Reasoning

The speed should be w/k which would give 2.42pi/0.0276pi = 87.68 cm/s

The max transverse velocity though...would that be when the trig function of cos is equal to 1 in this equation u = -wAcos(kx-wt) ?
 
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The speed is correct. Transverse speed is when dy/dt is a max. I think you are pretty much nailing that as well.
 
Apologies...still not quite sure...would I use the w and A i know multiplied with cos(kx-wt) to get my answer...so if cos(kx-wt)=1, then max transverse speed would be -wA?
 
If your displacement function is a sin wave, it's not possible that your velocity "function" be a constant of 87.68cm/s as you calculated. Your speed would be a function as well.

As you know, velocity is the change of displacement over time (i.e. dy/dt), so by differentiating the function for y, you will get a the velocity function (which in this case is a cosine function). That cosine function will allow you to calculate the speed at any location x and time t.

To calculate the maximum speed, we take a look at the velocity function. If you do your calculus right, the equation should look something like:

v = B cos (Cx + Dt)

There is an infinite possibility for the values of x and t, so we can assume that hthe cosine can return any value. The maximum value however of a cosine function is 1. So what does that tell you about your maximum velocity?
 
croyboy said:
Apologies...still not quite sure...would I use the w and A i know multiplied with cos(kx-wt) to get my answer...so if cos(kx-wt)=1, then max transverse speed would be -wA?

Yes. Except I'd call the maximum speed +wA. Why would you say minus?
 
Thanks much...makes a lot more sense...appreciate the help
 

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