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 344 m/5 piano A produces a wavelength of 0.763 m, while piano B produces a wavelength of 0.778 m. How much time separates successive beats?

Number______ Units_____

Answer 1

The time separation between the successive beats of the out-of-tune pianos is -0.028 s, indicating constructive interference.

Step-by-step explanation:

The time separation between successive beats can be calculated using the formula:

Time separation = 1 / (f2 - f1)

where f1 and f2 are the frequencies of the two waves.

To find the frequencies, we can use the formula:

f = v / λ

where f is the frequency, v is the speed of sound, and λ is the wavelength.

For piano A:

f1 = 344 / 0.763 = 449.93 Hz

For piano B:

f2 = 344 / 0.778 = 442.47 Hz

Substituting these values into the formula for time separation:

Time separation = 1 / (442.47 - 449.93) = -0.028 s

The negative sign indicates that the beats are in phase, i.e., they are constructive interference. Therefore, there is no time separation between the successive beats.

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