Solved: A musical interval of an octave corresponds to a

Solution 21DQ

Step 1:

Suppose, the initial frequency is,

Where, T - tension on the string

             - Linear mass density of the string

             L - length of the string

If we are raising the pitch to one octave, then, the frequency will be, 2

So, we can write,

Where, T’ - New tension on the string

Dividing both equations,

Squaring both sides, we get, 4 = T’ / T

Or, T’ = 4T

So, we have to apply a tension 4 times higher than the initial tension to raise the pitch to one octave.