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...
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?
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.