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 343 m/s, piano A produces a wavelength of 0.762 m, while piano B produces a wavelength of 0.781 m. How much time separates successive beats?

Answer 1

time period = 0.09 s

Step-by-step explanation:

given data

speed of sound v = 343 m/s

wavelength λ1 = 0.762 m

wavelength λ2 = 0.781 m

solution

we know that beat frequency is the difference between the frequency of the individual wave so

frequency f1 =
`(v)/(\lambda 1)` ..........................1

frequency f1 =
`(343)/(0.762)` = 450.13

and

frequency f2 =
`(343)/(0.781)` = 439.18

so beat frequency = f1 - f2

beat frequency = 450.13 - 439.18

beat frequency = 10.95 Hz

and

time period between successive beats is

time period =
`(1)/(f)`

time period =
`(1)/(10.95)`

time period = 0.09 s