Question

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 596 N to 538.00 N, what beat frequency is heard when the hammer strikes the two strings simultaneously

Answer 1

Answer:

5.4893 beats/sec

Explanation:

We have f_1 and f_2 as frequency of two waves

also we know that according to question

$f_2=f_1\sqrt{\frac{F'}{F} }$

F and F' are forces on the two frequencies.

therefore, beat frequency is

$f=f_1-f_2=f_1-f_1\sqrt{\frac{F'}{F} }\\=110(1-\sqrt{\frac{538}{596} })\\=5.4893 \text{ beats/sec}$