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Over [174]
4 years ago
15

Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observ

er stands at X, 4 m in front of one speaker. If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m
Physics
1 answer:
zhenek [66]4 years ago
8 0

Answer:

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is: <u>                  </u>

From above;  the distance between speaker  A and speaker B can be expressed as:

\sqrt{3^2 + 4^2 } \\ \\ =  \sqrt{9+16 } \\ \\ = \sqrt{25}  \\ \\ = 5 \ m

The path length difference  will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length  will be half the wavelength; which is

= \dfrac{1}{2}*4 \ m

= 2 m

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Learn more about The law of momentum conservation here: brainly.com/question/7538238

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