## A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an amplitude of 0.1

Question

A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an amplitude of 0.101 mm and a frequency of 71.0 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x,t) = sin(kx + ωt), what are (a) , (b) k, and (c) ω, and (d) the correct choice of sign in front of ω?

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22 mins 2021-07-22T09:52:25+00:00 1 Answers 0 views 0

Explanation:

Given that,

Mass per unit length is

μ = 4.87g/cm

μ=4.87g/cm × 1kg/1000g × 100cm/m

μ = 0.487kg/m

Tension

τ = 16.7N

Amplitude

A = 0.101mm

Frequency

f = 71 Hz

The wave is traveling in the negative direction

Given the wave form

y(x,t) = ym• Sin(kx + ωt)

A. Find ym?

ym is the amplitude of the waveform and it is given as

ym = A = 0.101mm

ym = 0.101mm

B. Find k?

k is the wavenumber and it can be determined using

k = 2π / λ

Then, we need to calculate the wavelength λ using

V = fλ

Then, λ = V/f

We have the frequency but we don’t have the velocity, then we need to calculate the velocity using

v = √(τ/μ)

v = √(16.7/0.487)

v = 34.29

v = 5.86 m/s

Then, we can know the wavelength

λ = V/f = 5.86 / 71

λ = 0.0825 m

So, we can know the wavenumber

k = 2π/λ

k = 2π / 0.0825