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

Answer:

Time period between the successive beats will be 0.1703 sec

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 $f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz$

Frequency of piano B $f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz$

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period $T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec$

So time period between the successive beats will be 0.1703 sec