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A commonly used physics experiment that examines resonances of transverse waves

ID: 1955264 • Letter: A

Question

A commonly used physics experiment that examines resonances of transverse waves on a string is shown below. A weight is attached to the end of a string draped over a pulley; the other end of the string is attached to a mechanical oscillator that moves up and down at a frequency f that remains fixed throughout the demonstration. The length L between the oscillator and the pulley is fixed, and the tension is equal to the gravitational force on the weight. For certain values of the tension, the string resonates. Assume the string does not stretch or shrink as the tension is varied. You are in charge of setting up this apparatus for a lecture demonstration.

(d) For your particular setup to fit onto the lecture table, you chose L = 1.10 m, f = 72.8 Hz, and = 0.794 g/m. Calculate how much tension is needed to produce each of the first three modes (standing waves) of the string.

w1 =  N w2 =  N w3 =  N

Explanation / Answer

Length L = 1.1 m Fundamental frequency f = 72.8 Hz Linear density = 0.794 x 10 -3 kg / m We know Fundamental frequency f = ( 1/2L) [ T / ] From this tension T = (2Lf) 2                               = 20.36 N i.e., W 1 = T               = 20.36 N Second harmonic frequency f ' = 2f                                                = 145.6 Hz We know f ' = ( 2/2L) [ T '/ ] From this f T Or ( f ' / f ) = [ T ' / T ] From this required tension T ' = T ( f ' / f ) 2                                                = 4 T                                                = 81.46 N i.e., W 2 = T '               = 81.46 N Third harmonic frequency f " = 3f                                             =  218.4 Hz We know f " = ( 3/2L) [ T "/ ] From this f T Or ( f " / f ) = [ T " / T ] From this required tension T " = T ( f " / f ) 2                                                = 9 T                                                = 183.24 N i.e., W 2 = T '               = 183.24 N Third harmonic frequency f " = 3f                                             =  218.4 Hz We know f " = ( 3/2L) [ T "/ ] From this f T Or ( f " / f ) = [ T " / T ] From this required tension T " = T ( f " / f ) 2                                                = 9 T                                                = 183.24 N i.e., W 2 = T '               = 183.24 N
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