Question

A note has a frequency of $128{\text{Hz}}$. Find the frequency of a note two octaves higher than it.
A) $256{\text{Hz}}$
B) ${\text{64Hz}}$
C) ${\text{32Hz}}$
D) $512{\text{Hz}}$

Answer 1

An octave is referred to as an interval between two notes. It can be between an upper note and a lower note where the frequency of the lower note is half of the upper note. It can also be between a lower note and an upper note where the frequency of the upper note is twice that of the lower note.

Complete step by step answer:
Step 1: List the parameters given in the problem.
Here a note having a frequency $f = 128{\text{Hz}}$ is given. We are to determine the frequency of another note which is two octaves higher than the given note.

Step 2: Express the frequency of the required note.
One octave higher corresponds to the frequency being twice that of the given note. So two octaves will correspond to the frequency of the required note to be four times that of the given note.
Let $f$’ be the frequency of the note two octaves higher than the given note.
Then we have $f' = 4f$ ------ (1)
Substituting $f = 128{\text{Hz}}$ in equation (1) we get, $f' = 4 \times 128 = 512{\text{Hz}}$.
Thus the frequency of that note which is two octaves higher than the given note is $f' = 512{\text{Hz}}$.

So the correct option is (D).

Additional information: When one says one octave higher it essentially means the same note but belonging in a higher section of the instrument. For example, in a piano, the keys to the right are higher than the ones to the left.

Note: Here it is mentioned that the frequency of the required note is two octaves higher than the given note. So we can automatically eliminate options B and C as the frequency of the required note must be greater than the given frequency $f = 128{\text{Hz}}$. An octave is essentially the interval between a musical pitch and another whose frequency is double or half of the other pitch.