(a) If a source emits sound of wavelengths 2. 64 m and 2. 72 m in air, 4 beats per second will be heard.
(b)The maximum intensity regions are apart by 90 m (approx).
Given,
Wavelengths of sound waves emitted by the source are respectively = 2.64 m and λ2 = 2.72 m
(a) As the measurement of the beats per second is equal to the frequency difference of the overlapping two sound waves,
Therefore, the frequency difference of the two waves is,
Δf =()
where and
are the frequencies of waves of wavelength λ1 and λ2
Now, speed of sound in air at temperature t = 20° C, is v = 343 m/s
∴ Δf = ()
⇒Δf =
⇒Δf =
⇒Δf ≅ 4
Hence 4 beats per second will be heard.
(b) When two waves overlap in the same phase in a region the intensity of that region becomes maximum due to constructive interference.
For constructive interference, the path difference is = nλ
Therefore the difference between the maximum intensity regions is given by,
d =
⇒d≅90 m
Hence the maximum intensity regions are apart by 90 m (approx).
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