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Tom walks around S1 in a circular path. The two loud speakers S1 (0, 0) and S2 (0,10m) are driven by the same oscillator are identical with adjustable frequencies and power 300-watt each. Assume sound intensity does not decay due to the distance and it is Io=kA2(A is the amplitude of a sound wave). When Tom is at point P (10, 0), he receives a maximum intensity of sound.
(a) What are the frequencies of the two speakers?
(b) What is the sound intensity and sound level (in dB) at point P?

Respuesta :

Answer:

Part a)

[tex]f = 82.8 Hz[/tex]

Part b)

[tex]I_{max} = 0.95 Watt/m^2[/tex]

[tex]L = 119.8 dB[/tex]

Explanation:

Path difference of two sounds reaching at the position of Tom is given as

[tex]\Delta L = L_1 - L_2[/tex]

here we know that

[tex]L_1 = 10\sqrt2[/tex]

[tex]L_2 = 10[/tex]

now we have

[tex]\Delta L = 10\sqrt2 - 10[/tex]

so we know that path difference must be equal to wavelength for maximum intensity of sound

so we have

[tex]\lambda = 10(\sqrt2 - 1)[/tex]

[tex]\lambda = 4.14 m[/tex]

now frequency of sound is given as

[tex]f = \frac{v}{\lambda}[/tex]

[tex]f = \frac{343}{4.14}[/tex]

[tex]f = 82.8 Hz[/tex]

Part b)

Intensity of source at position of Tom is given as

[tex]I = \frac{P}{4\pi r^2}[/tex]

so we have

[tex]I = \frac{300}{4\pi(10)^2}[/tex]

[tex]I = 0.24[/tex]

now due to constructive interference the maximum intensity is given as

[tex]I_{max} = 4I[/tex]

[tex]I_{max} = 0.95 Watt/m^2[/tex]

now sound level is given as

[tex]L = 10 Log\frac{I}{I_0}[/tex]

[tex]L = 10 Log\frac{0.95}{10^{-12}}[/tex]

[tex]L = 119.8 dB[/tex]

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