Suppose you pluck a string on a guitar and it produces the note A at a frequency of 440 Hz. Now, you press your finger down on the string against one of the frets, making this point the new end of the string. The newly shortened string has 4/5 the length of the full string. When you pluck the string, its frequency will be:
a) 350 Hz
b) 440 Hz
c) 490 Hz
d) 550 Hz
I think it is A because frequency and length of a string have an inverse relationship. Am I correct?
short string ----> higher frequency !!!
f = k/L
so (5/4) 440 = 550
Thank you!
You are welcome :)
Yes, you are correct! The frequency and length of a vibrating string indeed have an inverse relationship. When you press your finger down on the string against one of the frets, you effectively shorten the length of the string.
The frequency of a vibrating string can be calculated using the formula:
f = (1/2L) * √(T/μ),
where f is the frequency, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string.
Since the new length of the string is 4/5 times the original length, we can say that L_new = (4/5) * L.
Plugging this into the formula, we get:
f_new = (1/2 * L_new) * √(T/μ)
= (1/2 * (4/5) * L) * √(T/μ)
= (2/5 * L) * √(T/μ)
= (2/5) * (1/2 * L) * √(T/μ)
= (2/5) * f_original,
where f_original is the original frequency of the string before pressing the finger down on the fret.
Therefore, the frequency of the newly shortened string is (2/5) times the original frequency.
Since the original frequency is 440 Hz, the frequency of the newly shortened string will be:
f_new = (2/5) * 440 Hz
= 176 Hz.
So, you are incorrect in your answer choice. The correct answer is 176 Hz.