Low E on a guitar vibrates at 82.4 Hz and the guitar is 62.8 cm long. How long would the same string have to be to vibrate at high E 329.6 Hz. Assume that both strings vibrate at the fundamental frequency for the string.
To determine the length of the string required to vibrate at a certain frequency, we can use the formula for the fundamental frequency of a vibrating string:
f = (v/2L) * n
where:
- f is the frequency of vibration
- v is the speed of the wave (which is characteristic of the string material and tension)
- L is the length of the string
- n is a positive integer representing the harmonic order (in this case, we're considering the fundamental frequency, so n = 1)
From the given information:
- For the low E string, f = 82.4 Hz and L = 62.8 cm
- For the high E string, we need to find L when f = 329.6 Hz
Let's rearrange the formula to solve for L:
L = (v/2f) * n
For the low E string:
L₁ = (v/2f₁) * 1
For the high E string:
L₂ = (v/2f₂) * 1
The speed of the wave, v, is the same for both strings, so we can cancel it out:
L₂ = (1/2f₂) * f₁ * L₁
Now, substitute the values:
L₂ = (1/2 * 329.6 Hz) * (82.4 Hz) * (62.8 cm)
Calculating this value:
L₂ = 0.25 * 82.4 Hz * 62.8 cm
L₂ = 0.25 * 5153.12 cm
L₂ = 1288.28 cm
Therefore, the same string would need to be approximately 1288.28 cm long to vibrate at high E 329.6 Hz.
To find the length of the string required for the high E (329.6 Hz), we can use the formula for the fundamental frequency of a vibrating string:
f = (v/2L) * n
where:
f is the frequency of the string,
v is the velocity of the wave (which is constant for a given string),
L is the length of the string, and
n is a constant that depends on the vibrating mode of the string.
In this case, both strings are vibrating at their respective fundamental frequencies, so n = 1 for both.
Given that the low E string vibrates at 82.4 Hz and has a length of 62.8 cm, we can rearrange the formula to solve for L:
L = (v/2f) * n
For the low E string:
L₁ = (v/2f₁) * n = (v/2*82.4) * 1 = v/164.8
Now, to find the required length of the high E string (f₂ = 329.6 Hz), we can use the same formula:
L₂ = (v/2f₂) * n = v/([2*329.6]) * 1 = v/659.2
Comparing the two formulas, we have:
L₂ = (v/164.8) * [v/659.2]
L₂ = (v/v) * (1/4)
L₂ = 1/4
Therefore, the length of the high E string should be one-fourth (1/4) of the length of the low E string.
To calculate the length:
L₂ = (1/4) * 62.8 cm
L₂ = 15.7 cm
So, the length of the string required for the high E (329.6 Hz) is 15.7 cm.