Question

When one note is above another note by an octave, the frequency of the higher pitch note is double that of the lower pitch note. The word octave may make you think that there are 8 steps from the lower note to the one an octave above. But wait! How many steps are there on a piano from middle C to the next C above? Each key is a step. If each next key is a step, how many are there? Middle C is a white (W) key and the next note above is a black (B) key. Starting at middle C on the piano the sequence of key we go through to get to high C, is BWBWWBWBWBWW, the last key being high C. Count ‘em. That’s 12 steps. If the multiplying factor of the frequency for the next step is the same for each step, what is that multiplying frequency? All you need is given in this problem statement.

121/8

21/12

101/12

21/8

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to four times its length and make it vibrate in the fundamental frequency once again. The rubber band is made so that quadrupling its length quadruples the tension and reduces the mass per unit length by a factor of 4. The new frequency will be related to the old by a factor of

42.

V4 .

4.

1/4

1.

If I quadruple the mass per unit length of guitar string, its natural frequency changes by what factor?

42

1

4

Answer 1

Not sure about the first question, Here is second and third question

We know

f = v / 2L

where v = sqrt (T / u)

where T is tension and u is mass per unit length

tension is increased by 4 and mass per unit length is decreased by 4

so,

v = sqrt (4T / u/4)

v = sqrt ( 16T/u)

or

v = 4 sqrt (T/u)

so, we have

now,

f = v / 2*4L

f = 4 sqrt (T/u) / 4 * 2L

we can see that

f' = 4v/ 4*2L

f' = v/2L

so,

factor is 1,

---------------------------------------------------------------------------------

If we quadruple mass per unit length

v = sqrt (T/4u)

so,

f = sqrt (1/4) sqrt (v/u) / 2L

so,

factor is $1/\sqrt{4}$