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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.

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