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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?

Respuesta :

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 [tex]A\lambda _A=0.766m[/tex]

Wavelength of piano  [tex]B\lambda _B=0.776m[/tex]

So frequency of piano A [tex]f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz[/tex]

Frequency of piano B [tex]f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz[/tex]

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period [tex]T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec[/tex]

So time period between the successive beats will be 0.1703 sec

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