Two out of phase loudspeakers are some distance apart. A person stands 4.50 m from one speaker and 3.80 m from the other. What is the third lowest frequency at which the person will hear constructive interference? The speed of sound in air is 347 m/s. Hz

Answer :

Answer:

The third lowest frequency is 1489.27 Hz.

Explanation:

Given that,

Distance of person from one speaker= 4.50 m

Distance of person from the other = 3.80 m

The longest wavelength is the shortest frequency.

We need to calculate the difference between the speakers

The difference between the speakers d= 4.50-3.80 =0.7 m

We know that,

The longest wavelength is such that one wavelength

[tex]\lambda = d[/tex]

So, The third longest wavelength is

[tex]3\lambda=d[/tex]

Put the value of d

[tex]\lambda=\dfrac{0.7}{3}[/tex]

[tex]\lambda=0.233\ m[/tex]

We need to calculate the lowest frequency

Using formula of frequency

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

[tex]f=\dfrac{347}{0.233}[/tex]

[tex]f=1489.27\ Hz[/tex]

Hence, The third lowest frequency is 1489.27 Hz.

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