1. A thin taut string of mass 5.00 g is fixed at both ends and stretched...

1. A thin taut string of mass 5.00 g is fixed at both ends and stretched such that it has two adjacent harmonics of 525 Hz and 630 Hz. The speed of a traveling wave on the string is 168 m/s.

a.Determine which harmonic corresponds to the 630 Hz frequency.

b.Find the linear mass density of this string.

c.Find the tension in the string.

Solutions

Expert Solution

(a) Wavelength of harmonic of 525 Hz = ( 168 / 525 ) m = 0.32 m.

Wavelength of harmonic of 630 Hz = ( 168 / 630 ) m ~ 0.267 m.

Hence, difference in wavelength = ( 0.32 - 0.267 ) m = 0.053 m.

We know, difference between two successive wavelengths for a string fixed at both ends = / 2,

where, = Actual wavelength, in m.

Hence, / 2 = 0.053, or, = ( 0.053 x 2 ) m = 0.106 m.

Also, we know, the n^th harmonic has a wavelength n / 2, for a string fixed at both ends.

So, wavelength of harmonic of 630 Hz :

n / 2 = 0.267

or, n = 2 x 0.267 / 0.106

or, n = 5, taking the closest integer.

Hence, the 630 Hz frequency sound corresponds to the 5^th harmonic.

(b) Wavelength of fundamental frequency = / 2 = Length of the string, L, in m

Hence, L = 0.106 m = 10.6 cm.

Mass of the string = 5 g.

So, linear mass density of the string : m = 5 g / 10.6 cm ~ 0.47 g / cm.

Hence, tension T in the string is given by : v = ( T / m )

or, T = mv² = ( 0.47 x 168² ) N ~ 1.33 x 10⁴ N.

genius_generous answered 1 year ago

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