Question

A source emits sound of wavelengths 2. 64 m and 2. 72 m in air.

Answer 1

(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 =($f_{1} -f_{2}$)

where $f_{1}$ and $f_{2 }$ 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 = ($f_{1} -f_{2}$)

⇒Δf = $\frac{v}{\lambda1}-\frac{v}{\lambda2}$

⇒Δf =$\frac{343}{2.64} -\frac{343}{2.72\\}=3.82$

⇒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 = $\frac{v}{\Delta f} = \frac{343}{3.82\\}= 89.79$

⇒d≅90 m

Hence the maximum intensity regions are apart by 90 m (approx).

Learn more about constructive interference on

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