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A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an amplitude of 0.12 mm and a frequency of 100 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x, t) = ym sin(kx ± ωt), what are (a) ym, (b) k, (c) ω, and (d) the correct choice of sign in front of ω?

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Answer:

0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +

Explanation:

Given the wave equation of the form :

y(x, t) = ym sin(kx ± ωt)

Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m

Tension, T = 10 N

Amplitude, A = 0.12 mm

Frequency, F = 100 Hz

Comparing with the general wave equation :

y = Asin(kx ± ωt)

A = amplitude = ym = 0.12 mm

2.) k = 2π / λ

Recall :

v = fλ

v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472

λ = v/ f = 4.472 / 100 = 0.04472

Hence,

k = (2 * π) / 0.04472

k = 140.50 rad/m

3.) Angular frequency, ω

ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec

4.) sign is +ve

Direction of wave propagation as given is in the negative x axis

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