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Two speakers are placed 4.5m apart, playing a coherent 660 Hz tone. A listener is at a point 0.20 m from one speaker, on the line between the two speakers. The listener increases the frequency until the loudness they hear reaches a maximum. Calculate the first frequency at which this occurs. Take the speed of sound in air to be 344 ms-1.

Respuesta :

Answer:

The first frequency is 671 Hz.

Explanation:

Given that,

Frequency = 660 Hz

Distance from one speaker = 0.20 m

Distance between two speakers = 4.5 m

Speed of sound = 344 m/s

We need to calculate the path difference

Path difference = 4.3-0.20 =4.1 m

We need to calculate the frequency

Using formula of path difference

[tex]\phi=n\lambda[/tex]

[tex]4.1=n\times\dfrac{v}{f}[/tex]

[tex]f=\dfrac{344\times8}{4.1}[/tex]

[tex]f=671\ Hz[/tex]

Here, n = 8

Because, the frequency is less then 660 Hz at n = 7.

Hence, The first frequency is 671Hz.

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