Transverse Velocity of Wave, given position & time

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SUMMARY

The discussion centers on calculating the transverse velocity of a particle on a transverse sine wave given specific parameters. The wave's angular frequency (ω) is 125.66 rad/s, amplitude (A) is 2.50 mm, and wave number (k) is 3.49 rad/m. The participant initially calculated the transverse velocity (Vy) at x = 1.35 m and t = 0.0625 s as 2.07 * 10-4 m/s, which was later identified as incorrect due to a misunderstanding of the phase constant and potential rounding errors. The correct answer for the velocity at that position and time is 0 m/s, as the particle is at maximum displacement.

PREREQUISITES
  • Understanding of harmonic wave equations, specifically y(x,t) = Acos(kx - ωt)
  • Knowledge of angular frequency (ω) and wave number (k) in wave mechanics
  • Familiarity with the concept of phase constant (φ) in wave equations
  • Ability to perform trigonometric calculations involving sine and cosine functions
NEXT STEPS
  • Learn about the implications of phase constants in wave equations
  • Study the relationship between wave parameters and particle velocity in harmonic motion
  • Explore how to express wave parameters in terms of π for simplified calculations
  • Investigate common rounding errors in physics calculations and how to avoid them
USEFUL FOR

Students and educators in physics, particularly those studying wave mechanics and harmonic motion, as well as anyone involved in solving problems related to transverse waves and their properties.

Ryaners
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Can't see where I'm going wrong here - would greatly appreciate if anyone can point it out!

I've gotten the other parts of the question right, so I know that:
ω = 125.66 rad/s
A = 2.50 * 10-3m
k = 3.49 rad/m
The wave is moving in the +x direction.

The general equation for the position of a particle on the wave is:
y(x,t) = Acos(kx - ωt)

So this means (please correct me if I'm wrong) that the transverse velocity of a particle on the wave is:
Vy(x,t) = ωAsin(kx - ωt) = (125.66)((2.50 * 10-3m)sin(3.49x - 125.66t)

I'm asked to find Vy at x = 1.35m, t = 0.0625s.
For this I got 2.07*10-4m/s, but I'm told this is incorrect. Where am I going wrong?
 
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Hello. Check to see if you have the overall correct sign for your answer.

The most general equation for a harmonic wave traveling in the positive x direction is y = Acos(kx-ωt+φ) where φ is a phase constant. Do you know if you are supposed to assume that the phase constant is zero in this problem? It's always a good idea to state the problem word-for-word so that we can be sure that we have all of the information.
 
Thanks for the response TSny. Here's the full text of the question:

"A transverse sine wave with an amplitude of
2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time t=0 the left end of the string has its maximum upward displacement."

I had to 'give up' on the question & the answer given was v = 0m/s - I could then see that the part of the question immediately before - which asked for the displacement of a particle at the same position & time - was exactly equal to the amplitude, so I probably should have realized that the velocity should be 0 there. I'm still puzzled as to why my calculation didn't give that answer though - rounding error, I suppose?
 
OK. The information given in the problem implies that the phase constant φ is zero. So, you have the correct mathematical form for the wave. Yes, I think your answer differs from zero due to round-off error. Try working the problem by expressing k and ω in terms of ##\pi##. Then when you evaluate the argument of the cosine function for the specific values of x and t given, you can get the argument in terms of ##\pi##.

Anyway, your method was correct. :smile:
 
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Thank you :) I'll give that a go when I've gotten through all of my other assignments...
 

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