Transverse Wave speed and acceleration

AI Thread Summary
The discussion centers on calculating the speed and acceleration of a transverse wave described by the function y = 0.25 sin[(π/2) + 4πt]. To find the wave speed at t = 2 seconds, the relationship between angular frequency and wave number is utilized, with the formula v = ω/k. The participants clarify that the wave function lacks a spatial coordinate, which is crucial for proper differentiation. The confusion about constants in the wave equation is addressed, leading to a clearer understanding of the derivatives needed for solving the problem. Overall, the conversation emphasizes the importance of correctly interpreting wave functions in physics calculations.
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Homework Statement



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

y=.25(meters) sin[\frac{\pi(4)}{8} + \pi4t]

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

b.) Find the acceleration at t= 2sec

Homework Equations



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

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

0.25 \sin{\left(\frac{4\pi}{8}+4\pi t\right)}

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

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

where k is the wave number and \omega is the angular frequency. Therefore the velocity is \frac{dx}{dt}=\frac{\omega}{k}
 
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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