Transverse Wave speed and acceleration

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Homework Help Overview

The discussion revolves around a transverse wave described by a mathematical function, with specific questions regarding the wave's speed and acceleration at a given time. The subject area is wave mechanics, focusing on the properties of transverse waves on a string.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the speed and acceleration of the wave by taking derivatives but expresses confusion about the process and whether it is necessary. Some participants question the formulation of the wave function and its implications regarding spatial coordinates.

Discussion Status

Participants are engaging in clarifying the wave function and its components. One participant provides a mathematical insight regarding the relationship between wave speed, angular frequency, and wave number, which seems to help the original poster gain clarity on the constants involved.

Contextual Notes

There is a note about the original wave function lacking a spatial coordinate, which raises questions about its interpretation. The original poster also mentions uncertainty about the derivative process and the correct interpretation of constants in the equation.

ironlee
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Homework Statement



A Transverse wave on a string is described by this function :

y=.25(meters) sin[[tex]\frac{\pi(4)}{8}[/tex] + [tex]\pi[/tex]4t]

a.) Find the speed of the wave at t= 2sec

b.) Find the acceleration at t= 2sec

Homework Equations



y=.25 sin[[tex]\frac{\pi(4)}{8}[/tex] + 4[tex]\pi[/tex]t]

The Attempt at a Solution



I tried taking the derivative with respect to t, but man I can't figure it out for the love of god and I have no idea what I'm doing wrong. I doubted myself so much I don't even know if I need to take the derivative.

P.S. at the end of the equation its 4 * pi * t (not 4 to the power of Pi) and its .25 meters
 
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Are you sure you meant to write

[tex]0.25 \sin{\left(\frac{4\pi}{8}+4\pi t\right)}[/tex]

because there isn't any spatial coordinate in that wave function.

Either way, for a transverse wave, the argument of the function is constant, so

[tex]\frac{d}{dt}\left(kx-\omega t\right)=k\frac{dx}{dt}-\omega=\frac{d}{dt}C=0[/tex]

where [itex]k[/itex] is the wave number and [itex]\omega[/itex] is the angular frequency. Therefore the velocity is [tex]\frac{dx}{dt}=\frac{\omega}{k}[/tex]
 
Thank you so much! That'll be it, makes sense and yeah, that is what I meant to write. My problem was I couldn't remember which were constant. Thanks again!
 
Oh ok, great, I thought that's what you meant. Glad to help.
 

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