A string of mass 5 grams in stretched between two poles 2.5 m apart. The 5th har

Question

A string of mass 5 grams in stretched between two poles 2.5 m apart. The 5th harmonic has a frequency of 300 Hz. The amplitude at the antinodes is 4 mm.

a) What is the tension in the string?

I answered 225,000 N. I feel like that's too large though and would like a second opinion.

b) Write down an equation for the displacement of the string as a function of position (x) and time (t).

I answered: y(x,t) = 0.004 sin (1.26x - 377t). I found values for k and and plugged them in. Not too confident in this answer as well and need a second opinion.

c) What is the maximum transverse speed of an element of the string at an antinode? What is the transverse displacement of the string element when the transverse speed is the maximum? Justify your answer.

I was not able to answer the second part and the first part is VERY iffy. I used the formula um=ym* and got um (or what i Imagine to be the maximum transverse speed) to be 1.5 m/s. It's too small in my opinion, so I'm not sure.

Explanation / Answer

a) L = 2.5 m

wavelength = 2L/n = 2*2.5/5 = 1.00 meter

speed of wave = wavelength * freq = 1.00 * 300 = 300 m/s

v^2 = tension / (mass/length)

300^2 = T / (0.005/2.5) note: mass must be in kg (watch your units!)

T = 90000 * .005 / 2.5 = 180 newtons

b) you have to use the equation for a standing wave, which is y = A sinkx coswt

in this case k = 2pi/wavelength = 6.28 (in standard units)

w = 2pif = 2*3.1416*300 = 1885 (in st units)

and A = 0.004 (st units)

so the equation is y = 0.004 sin(6.28x) cos(1885t)

c) "transverse" refers to the oscillatory motion, so we know

v max = A w = 0.004 * 1885 = 7.54 m/s

where does it occur? when y = 0, i.e. at the midpoint of the oscillation (just like with a simple spring and mass oscillator).

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