A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to form a standing
wave with three antinodes when driven at a frequency of 420 Hz.
A) What is the frequency of the fifth harmonic of this string?
B) What is the tension in the string?
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Note that number of nodes and anti-nodes on a vibrating string determine the nth-harmonics. The fundamental frequency or first harmonic has 2 nodes and 1 anti node. The second harmonic has 3 nodes and 2 anti-nodes. The third harmonic has 4 nodes and 3 anti-nodes. As the nth-harmonic increases by 1, the number of nodes and anti-nodes also increases by 1.
Moreover, frequency of nth-harmonics are all multiples of the fundamental frequency. So,
`f_n = n*f_1`
where `n` - is the harmonic number
`f_1` - is the fundamental frequency
`f_n` - is the frequency of nth-harmonic
In the problem, the given frequency 420 Hz has 3 anti-nodes. Hence, it is the 3rd harmonic.
(A) f_5 = ?
To solve for the frequency of the fifth harmonic, fundamental frequency must be determined.
`f_n = n *f_1`
Substitute the third harmonic 420 Hz and the value of n (n=3) .
`420/3 = f_1`
The fundamental frequency is 140 Hz. Then, substitute this value to the formula to determine `f_5` .
`f_5 = nf_1`
`f_5 = 5 (140)`
`f_5 = 700`
Hence, the frequency of the fifth harmonic is 700 Hz.
(B) Tension in the string = ?
To solve for the tension in the string, speed of the wave must be determined.
`v = f_1lambda`
Note that the wavelength of a fundamental frequency is twice the length of the string. The given length of the string is 60 cm (0.6m).
`v = f_1(2L)`
Then, substitute value of `f_1` (140Hz) and L (0.60m).
`v= 140(2)(0.60) = 168 `
The speed of the wave is 168 m/s. Note that `f_3` , `f_5` and other nth-harmonics in the string has the same speed 168 m/s.
Then, use the formula of the speed of the wave in a stretch string, which is:
`v = sqrt(T/rho)`
where T - tension in the string (N)
`rho` - linear density (kg/m)
Note that the given density is 2 g/m. It still has to be converted to kg/m.
`168 = sqrt(T/(2/1000))`
`168 = sqrt(T/0.002)`
Square both side to remove the square root.
`168^2 = T/0.002`
`28224 = T/0.002`
`28224(0.002) = T`
The tenison in the string is 56.45N
We know that a fixed string like this have the wave length ` ` as following depending on the harmonics.
` ` = 2/n(L) where L is the length of the string and n is the number of harmony.
If you have three antinodes then;
(lambda/2)*3 = L
lambda = 2L/3
By applying V= f(lambda) to the first situation;
V = 420*2*0.6/3 = 168 m/s
For the fifth harmonic;
lambda = 2L/5 = 2*0.6/5 = 0.24m
By using equation V= f(lambda);
V = f_5*0.24
Since we have same string V=168m/s
So f_5 = 700 Hz
Fifth harmonic is 700 Hz
Since we can only answer one question as adviced by the editors i will omit the second part.
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