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Two loudspeakers are located 3.0 m apart on the stage of an auditorium. A listener at point P is seated 19.0 m from one speaker and 15.0 m from the other. A signal generator drives the speakers in phase with the same amplitude and frequency. The wave amplitude at P due to each speaker alone is A. The frequency is then varied between 30 Hz and 400 Hz. The speed of sound is 343 m/s. At what frequency or frequencies will the listener at P hear a maximum intensity?

Answer :

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Answer:

The frequencies that the listener at P will hear a maximum intensity are: 85.75 Hz, 171.5 Hz, 257.25 Hz and 343 Hz.

Explanation:

Path Difference Δx is given as: nλ

where λ = [tex]\frac{v}{f}[/tex]

Δx can be re-written as: n×[tex]\frac{v}{f}[/tex]

where;

n = integer

v = speed of sound = 343 m/s

f = frequency

Δx  = 19.0 m - 15.0 m

Δx  = 4.0 m

[tex]4.0 m =[/tex] [tex]\frac{n*343}{f}[/tex]

f = [tex]\frac{n*343}{4}[/tex]

f = n × 85.75 Hz

Now;  n = 1, 2, 3, 4 ; the frequencies that the listener at P will hear the maximum intensity will be

when n= 1

f = 1 × 85.75 Hz

f = 85.75 Hz

when n= 2

f = 2 × 85.75 Hz

f = 171.5 Hz

when n= 3

f = 3  × 85.75 Hz

f = 257.25 Hz

when n= 4

f = 4 × 85.75 Hz

f = 343 Hz

∴ the frequencies that the listener at P will hear the maximum intensity will be : 85.75 Hz, 171.5 Hz, 257.25 Hz and 343 Hz for n =1,2,3, and 4 respectively.

The maximum intensity, for two loudspeakers, which will the listener hear is 85.75, 171.5, 257.25, 343 Hz.... for n= 1,2,3,4... respectively.

What is frequency?

Frequency of a wave is the number of waves, which passed thorough a particular point at a unit time.

It can be given as,

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

Here, (v) is the speed of the wave and [tex]\lambda[/tex] is the wavelength.

The difference in the distance can be given as,

[tex]\Delta d=n\lambda[/tex]

Here, (n) is the integer number of wavelength. The frequency can be given as,

[tex]f=\dfrac{nv}{\Delta d}[/tex]

A listener at point P is seated 19.0 m from one speaker and 15.0 m from the other. Thus, the difference of distance is,

[tex]\Delta d=19-15\\\Delta d=4\rm m\\[/tex]

The speed of sound is 343 m/s. Put the values in above formula as,

[tex]f=\dfrac{n\times343}{4}[/tex]

Put value of n as 1 in the above formula, we get,

[tex]f=\dfrac{1\times343}{4}\\f=85.75\rm Hz[/tex]

Put the values of n as 2,3,4... to find the difference values of the maximum intensity.

Thus, the maximum intensity, for two loudspeakers, which will the listener hear is 85.75, 171.5, 257.25, 343 Hz.... for n= 1,2,3,4... respectively.

Learn more about the frequency here;

https://brainly.com/question/1436295

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