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 349 m/s, piano A produces a wavelength of 0.766 m, while piano B produces a wavelength of 0.776 m. How much time separates successive beats?

Answer 1

Time period between the successive beats will be 0.1703 sec

Step-by-step explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano
`A\lambda _A=0.766m`

Wavelength of piano
`B\lambda _B=0.776m`

So frequency of piano A
`f_1=(v)/(\lambda _1)=(349)/(0.766)=455.61Hz`

Frequency of piano B
`f_2=(v)/(\lambda _1)=(349)/(0.776)=449.74Hz`

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period
`T=(1)/(f)=(1)/(5.87)=0.1703sec`

So time period between the successive beats will be 0.1703 sec