Transverse Wave speed and acceleration

Click For Summary
SUMMARY

The discussion focuses on calculating the speed and acceleration of a transverse wave represented by the function y = 0.25 sin(π/2 + 4πt). The wave speed is determined using the formula v = ω/k, where ω is the angular frequency and k is the wave number. At t = 2 seconds, the wave speed and acceleration can be derived by taking the appropriate derivatives of the wave function. The correct interpretation of the wave function is crucial for accurate calculations.

PREREQUISITES
  • Understanding of transverse wave equations
  • Knowledge of derivatives in calculus
  • Familiarity with angular frequency and wave number concepts
  • Ability to interpret trigonometric functions in wave mechanics
NEXT STEPS
  • Learn how to derive wave speed from wave equations
  • Study the relationship between angular frequency and wave number
  • Explore the concept of acceleration in wave motion
  • Practice solving problems involving derivatives of trigonometric functions
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, as well as educators and tutors looking for examples of transverse wave calculations.

ironlee
Messages
2
Reaction score
0

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
 
Physics news on Phys.org
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.
 

Similar threads

Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
632
Replies
29
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K