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

Answer:

time period = 0.09 s

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 = $\frac{v}{\lambda 1}$   ..........................1

frequency f1 = $\frac{343}{0.762}$ = 450.13

and

frequency f2 =  $\frac{343}{0.781}$   = 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 = $\frac{1}{f}$  

time period = $\frac{1}{10.95}$

time period = 0.09 s