For a demonstration of standing waves a string with length 1.4 m is attached to

Question

For a demonstration of standing waves a string with length 1.4 m is attached to an oscillator operating at 17 Hz. The other end of the string passes over a pulley to a hanger where various masses can be attached to vary the tension in the string. Transverse standing waves are set up in the string with a node at the pulley and an anti-node at the oscillator. The mass per unit length of the string is 2 g/m. (a) What must the tension be for the string to vibrate with its lowest frequency? (b) What tensions are needed for the three next higher frequencies? T2 = N (largest value) T3 = N T4 = N (smallest value)

Explanation / Answer

L = length = 1.4 m

m = mass per unit length = 2 g/m = 0.002 kg/m

f = fundamental frequency = 17 Hz

T = Tension

fundamental frequency is given as

f = (1/(2L)) sqrt(T/m)

17 = (1/(2 x 1.4)) sqrt(T/0.002)

T = 4.53 N

for f2 = 2 f = 2 x 17 = 34 Hz

f2 = (1/(2L)) sqrt(T/m)

34 = (1/(2 x 1.4)) sqrt(T/0.002)

T = 18.12 N

for f3 = 3 f = 3 x 17 = 51 Hz

f3 = (1/(2L)) sqrt(T/m)

51 = (1/(2 x 1.4)) sqrt(T/0.002)

T = 40.78 N

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