I take a violin and make an exact copy of it, except that it is bigger. The strings are identical except for the length; they have the same material and the same tension. If the new violin is 2.30 times the size of the original, at what frequency would the string that was previously the A4 string (that is 440 Hz on a regular violin) oscillate? Use units of "Hz."
First of all "an exact copy" is just that = neither larger nor smaller.
Sra
To find the frequency at which the string would oscillate on the larger violin, you need to consider the relationship between the length of the string and its frequency of oscillation.
The frequency of oscillation of a string is inversely proportional to its length. This means that as the length of the string increases, the frequency decreases, and vice versa, given that all other factors remain constant.
In this case, the larger violin is 2.30 times the size of the original violin. Since we want to find the frequency of the string, which previously oscillated at 440 Hz on the original violin, we can set up a proportion to solve for the new frequency:
(original length) / (new length) = (original frequency) / (new frequency)
Let's assign some variables to the quantities we know:
- l1: original length of the string
- l2: new length of the string
- f1: original frequency (440 Hz)
- f2: new frequency (what we're trying to find)
Using the proportion, we have:
l1 / l2 = f1 / f2
Since the original frequency (f1) is 440 Hz and the size of the new violin is 2.30 times the size of the original, we can substitute these values into the equation:
l1 / (2.30 * l1) = 440 Hz / f2
Simplifying the equation:
1 / 2.30 = 440 Hz / f2
To solve for f2 (the frequency of the string on the larger violin), we can rearrange the equation:
f2 = (440 Hz) * (2.30 / 1)
Evaluating the expression:
f2 = 440 Hz * 2.30
Calculating the result:
f2 = 1012 Hz
Therefore, the string that was previously the A4 string (440 Hz) on the regular violin would oscillate at approximately 1012 Hz on the larger violin.