Question

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 344 m/s, piano A produces a wavelength of 0.770 m, while piano B produces a wavelength of 0.780 m. How much time separates successive beats?

Answer 1

Please note that two pianos when sounded together produce same note, but out of tune. This is called beating.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of two waves.

f(beat) = | f2 - f1|

Here, it is given that when speed of sound v = 344 m/s
lamda2 = 0.770

So, f2 = v / lemda2 = 344 / 0.770 = 447 Hz

And, lamda1 = 0.780

So, f1 = 344 / 0.780 = 441 Hz

So, f(beat) = | f2 - f1| = 447 - 441 = 6.0 Hz

Therefore, the time interval between = 1/f(beat) = 1/6.0 = 0.167 sec.