Transverse Wave Velocity/Acceleration

  • #1
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Homework Statement


The left-hand end of a long horizontal stretched cord oscillates transversely in SHM with frequency 270 Hz and amplitude 2.4 cm . The cord is under a tension of 90 N and has a linear density 0.08 kg/m . At t=0, the end of the cord has an upward displacement of 2.1 cm and is falling. Consider the point x = 1.30 m on the cord.
(solved A and B)
C. What is its velocity at t = 2.2 s ?
D. What is its acceleration at t = 2.2 s ?

Homework Equations


wave velocity v = √(T / μ)
wavelength λ = v/f
wave number k = 2π/λ
ω= 2πf
D(x,t) = Asin( kx - ωt + ∅) for transverse wave travelling to the right
Velocity = -Aωcos(kx - ωt + ∅)
Acceleration = -Aω*ωsin( kx - ωt + ∅)

The Attempt at a Solution



v = √(T / μ) = √(90/0.08) = 33.54 m/s (rounds to 34 m/s)
λ = v/f = 33.541/ 270 = 0.1242 m
k = 2π/λ = 2π/0.1242 = 50.578 1/m (rounds to 51 1/m)
ω= 2πf = 2π*270 = 1696 1/s (rounds to 1700 1/s)

Plugging things in, D(x,t) = 0.024 sin(51x -1700t + ∅)

(Therefore, part A max velocity is Aω = 0.024 m * 1700 1/s = 40.8 m/s (rounds to 41 m/s)
(Therefore, part B max acceleration is Aω*ω = 0.024*1700*1700 = 69360 m/s^2 (rounds 6.9 *10^4 m/s^2 )

At time t=0, amplitude is 0.021 m, so 0.021 = 0.024 sin(∅), ∅ = 1.06

Full equation: D(x,t) = 0.024 sin(51x -1700t + 1.06)
Velocity (1.30 m, 2.2 s) = -0.024*1700 cos(51*1.3 - 1700(2.2) + 1.06) = 40 m/s
Acceleration (1.30 m, 2.2 s) = -0.024*1700*1700 sin(51*1.3 - 1700(2.2) + 1.06)

Mastering physics says this is wrong (I also tried 41 m/s, thinking maybe there were rounding errors). Given that I got parts A and B correct, I can assume my A and ω are correct, but I can't find anything wrong with my k or ∅. The units work, it's a reasonable order of magnitude given the derived max velocity. I tried entering the equations with the values in terms of their formulas in wolfram alpha, and still wrong! I can't figure out what's wrong! I haven't tried entering my acceleration because my velocity is wrong.
 
Last edited:

Answers and Replies

  • #2
TSny
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Welcome to PF!
Looks like you plugged in x = 1.03 m instead of 1.30 m.
 
  • #3
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I wish! That's actually just an error in transcribing my work... Edited now, thanks
 
  • #4
TSny
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With x = 1.30 m I do not get your answer for v.
 
  • #5
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With x = 1.30 m I do not get your answer for v.

Yeah, that's where the 40.5 m/s (rounded to 41 m/s) comes in, using wolfram alpha. Are you getting something other than that?
 
  • #6
TSny
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It does seem to be differences in rounding the numbers. If I keep several extra digits in k, ω, and Φ, I get v to be about 27 m/s. It's peculiar that rounding makes this much difference.
 
  • #7
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It does seem to be differences in rounding the numbers. If I keep several extra digits in k, ω, and Φ, I get v to be about 27 m/s. It's peculiar that rounding makes this much difference.

Well that does seem to be the correct answer. How bizarre. Thanks much! Could I ask how many you kept, such that acceleration will be correct?
 
  • #8
TSny
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I let Mathematica do the calculations. It keeps quite a lot of digits! But it looks like things settle down if you keep about 6 digits.
 
  • #9
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I let Mathematica do the calculations. It keeps quite a lot of digits! But it looks like things settle down if you keep about 6 digits.

Thank you very much! All correct now. At least my theory was right, I guess...
 
  • #10
TSny
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It seems the peculiar behavior is due to the fact that for t = 2.2 s and x = 1.3 m, the argument of the cosine function for the velocity is quite large, about 3665 radians. A 0.1 % change in this argument amounts to almost 4 radians. Thus, the cosine will change a lot with just a small percent change in the argument.
 
  • #11
TSny
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At least my theory was right, I guess...
Yes, your work looks good.
 

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