There are 7 notes of "A" on a standard piano. When the first A is struck, the string associated with the key vibrates 55 times per second. The next A above the given A vibrates twice as fast. An exponential function with a base of two is used to determine the frequency of the 11 notes between the two A's. Find this function.
I have no clue what they mean or how to determine this, it is a question the teacher gave us and not in the book.
If you are already familiar with the piano keyboard, you'll find the following explanation not necessary. I include it just in case.
It is a little difficult to visualize without a keyboard in front of you. Here's a link to an image to which you can refer.
http://www.mouthmusic.com/pianokey.htm
The standard piano keyboard begins with an A key (A0, 27.5 Hz) and finishes with a C-key. There are 12 keys to an octave, named C,D,E,F,G,A,B (white keys) and C#,D#,F#,G#,A# (black keys to the right of the lettered key, pronounced C-sharp, D-sharp, etc.).
As your teacher said, the frequency is doubled as we progress through octaves to the right. For example, the leftmost A-key has a frequency of 27.5, and the next A-key to its right has a frequency of 55, and so on.
What you teacher stated is that the frequencies of all the (11) keys between the two A's are scaled geometrically.
If we denote f(A0) the frequency of the first A-key, and f(A1) the frequency of the next A-key, then f(A1)/f(A0)=2.
Since they are in geometric progression, then
f(A1)/f(G#) = f(G$)/f(G) = f(G)/f(F#) = f(F#)/f(F) .... = f(C)/f(B) = f(B)/f(A#) = f(A#)/f(A) = Ratio, R
He would like to see a calculation of all the intermediate frequencies between keys A1 (55 Hz) and A2 (110 Hz).
There are 12 spaces between the two keys. So f(A1)*R^12=f(A2), or
R^12=f(A2)/f(A1)=2
R=2^(1/12)=1.0595 approximately.
That is to say,
f(A#)=f(A1)*1.0595=58.27 Hz
f(B)=f(A#)*1.0595=61.74 Hz
f(C)=f(B)*1.0595 = 65.41 Hz
... and so on.
Post your calculations for a check if you wish.
To find the exponential function that represents the frequency of the 11 notes between the two A's, we need to understand a few concepts.
First, we know that the frequency doubles every time we go up one note. So, if the first A vibrates at 55 times per second, the next A above it will vibrate at twice that frequency, which is 55 * 2 = 110 times per second.
Let's assign the first A to the frequency f, which is 55 times per second. Therefore, the next A would have the frequency 2f, and the A after that would have the frequency 4f, and so on.
Now, for the 11 notes between the two A's, we need to find the number of times the frequency doubles. Since we have 11 notes, we need to double the frequency 11 times.
Using the concept of exponential growth, we can express the frequency of the nth note in terms of the first A's frequency, f, and the number of times it doubles, n.
The formula for exponential growth is:
y = a * (r)^x
where:
- y is the final value (frequency in this case)
- a is the initial value (f)
- r is the growth factor (2, as the frequency doubles every time)
- x is the number of times the growth factor is applied (n)
Applying this formula to find the frequency, we have:
frequency = f * (2)^n
Now, since we want to find the exponential function, let's assign the function as f(x):
f(x) = f * (2)^x
This is the exponential function that represents the frequency of the 11 notes between the two A's.
Note: In this context, x represents the number of notes (0 for the first A, 1 for the next note, 2 for the note after that, and so on), and f(x) represents the frequency of the corresponding note.
To find the exponential function that determines the frequency of the 11 notes between the two A's, let's break down the problem step by step.
1. We know that the first A vibrates at a frequency of 55 times per second.
2. The next A above vibrates twice as fast as the given A. This means its frequency is 55 * 2 = 110 times per second.
3. There are 11 notes between these two A's. Since we have a total of 7 A notes on a piano, we need to distribute the frequency ratios evenly across these 11 notes.
4. We can start by finding the common ratio of the exponential function. The common ratio is found by taking the ratio of the frequency of the next A (110) to the frequency of the current A (55). So, the common ratio = 110 / 55 = 2.
5. Now, we need to determine the base of the exponential function. Since we have a common ratio of 2, we can deduce that the base of the exponential function is 2.
6. The general form of an exponential function is given by y = a * b^x, where 'a' is the starting value, 'b' is the base, and 'x' is the exponent.
7. We know that the starting value (frequency of the first A) is 55, and we determined that the base is 2.
Putting it all together, the exponential function to determine the frequency of the 11 notes between the two A's can be written as: y = 55 * 2^x, where 'x' represents the number of notes away from the first A.
Now, you can substitute different 'x' values (ranging from 1 to 11) to find the respective frequencies of the notes between the two A's using this exponential function.