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Two loudspeakers, 4.0m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.35m. What is the frequency of the sound? If the frequency is then increased while you remain 0.35m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

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Answer:

Explanation:

Given

Distance between loud speaker is [tex]d=4\ m[/tex]

For destructive interference phase difference is given by

[tex]\Delta \phi =\frac{\lambda }{2}[/tex]

Distance of observer from one speaker

[tex]x_1=0.5L+0.35[/tex]

Distance of observer from second speaker

[tex]x_2=0.5L-0.35[/tex]

Phase difference

[tex]\Delta \phi =x_1-x_2=(0.5L+0.35)-(0.5L-0.35)[/tex]

[tex]\Delta \phi =0.7\ m[/tex]

therefore the wavelength of wave is

[tex]\Delta \phi =\frac{\lambda }{2}[/tex]

[tex]\lambda =0.7\times 2=1.4\ m[/tex]

frequency corresponding to this wavelength

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

where v=velocity of sound

[tex]=\frac{340}{1.4}=242.84\ Hz[/tex]    

For maximum intensity of sound

Phase difference [tex]\Delta \phi =\lambda[/tex]

[tex]\lambda =0.7\ m[/tex]

Frequency corresponding to this wavelength is [tex]f=\frac{340}{0.7}=485.71\ Hz[/tex]                    

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