# 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?

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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

Note:

Since we can only answer one question as adviced by the editors i will omit the second part.

Sources:

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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= 3f_1`

`420/3 = f_1`

`140=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).

So,

`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`

`56.45=T`

The tenison in the string is 56.45N

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