A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60cm apart. The string is observed to form a standing wave with three antinodes when driven at a frequency of 420 Hz. What are (a) the frequency of the fifth harmonic of this string and (b) the tension in the string?

Respuesta :

(a) 700 Hz

For standing waves on a string, the number of antinodes (n) corresponds to the order of the harmonic. So, three antinodes corresponds to the third harmonic. Also, the frequency of the nth-harmonic is the nth-integer multiple of the fundamental frequency, so we have:

[tex]f_3 = 3 f_1 = 420 Hz[/tex]

where [tex]f_1[/tex] is the fundamental frequency. Solving for f1, we find

[tex]f_1 = \frac{420 Hz}{3}=140 Hz[/tex]

And so now we can find the frequency of the 5th-harmonic:

[tex]f_5 = 5 f_1 = 5 (140 Hz)=700 Hz[/tex]

(b) 56.4 N

The fundamental frequency of a string is given by:

[tex]f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}[/tex]

where we have:

L = 60 cm = 0.60 m is the length of the string

[tex]\mu = 2.0 g/m = 0.002 kg/m[/tex] is the linear density

T = ? is the tension in the string

Solving the formula for T and using the fundamental frequency, f1=140 Hz, we find

[tex]T=\mu (2Lf_1)^2=(0.002 kg/m)(2(0.60 m)(140 Hz))^2=56.4 N[/tex]