Two consecutive harmonics on a stiff string with linear mass density = 0.100 !\"

Question

Two consecutive harmonics on a stiff string with linear mass density = 0.100 !" ! and under tension of 7290 N are 1350 Hz and 1530 Hz. i. Determine the speed of the wave on the string. [2 points] ii. Determine if sting is fixed (clamped) at one end or at both ends. Support your answer with calculations.[4 points] iii. If the string was vibrating next to a tube opened at both ends, filled with air (vSound=343 m/s), what would have to be the length of the tube that would make f=1350 Hz its fundamental frequency? What would be the next harmonic on the string that will excite the vibrations in the tube?

Explanation / Answer

fundamental frequency f = (1/2L) sqrt(T/u)

also velocity V^2 = T/u

where T is tension = 7290 N

u is linear mass density = 0. 1 kg/m

so here

Speed V^2 = 7290/(0.1)

V = sqrt(7290/(0.1) )

V = 270 m/s

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fundamental frequency = 1530-1350 = 180 Hz

No. of Harmonic n1 = 1350/180 = 7.5

also n2 = 1530/180 = 8.5

since this is not an integral no. here string must have connected closed at one end

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fundamental frequency f = V/2L

2L = V/f = 343/1350

L = 12.7 cm

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f2 = 2 * 343/(2 *0.127)

f2 = 2700 Hz

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