A student uses a 2.00-m-long steel string with a diameter of 0.90 mm for a stand

Question

A student uses a 2.00-m-long steel string with a diameter of 0.90 mm for a standing wave experiment. The tension on the string is tweaked so that the second harmonic of this string vibrates at 28.5 Hz.

(steel = 7.8 x 10^3 kg/m^3)

1. Calculate the tension the string is under.

Express your answer using two significant figures.

FT = _________N

2. Calculate the first harmonic frequency for this string.

Express your answer using three significant figures.

f1=__________Hz

3. If you wanted to increase the first harmonic frequency by 50%, what would be the tension in the string?

Express your answer using two significant figures.

FT1= _____________N

Explanation / Answer

linear density. P*pir^2=5e-3(kg/m). second harmonic at f=28.5 so that v=L/T=L*f=2*28.5=57(m/s). v=sqrt(F/5e-3) so F=16.2(N). ------- we have that lamda=2L/n. lamda=v*T=v/f. so v/f=2L/n. so f/n=const. so first harmonic. 28.5/2=f/1 so f=14.25(Hz) ------------ f=14.25*1.5 so velocity of wave. 2*2*14.25*1.5=85.5 so Ft=37(N)

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