Anonymous asked in 科學其他:科學 · 1 decade ago

大學普物問題!! Wave Optics!

A speaker, which emits a 200-Hz signal is 8 m from a microphone.

They are equidisant from a wall. What is the minimum distance to the

wall such that the sound that reaches the microphone directly, and

that which is reflected off the wall, interfere constructively.

Take the speed of sound to be 340m/s.

(There is no phase change on reflection.)

1 Answer

  • Anonymous
    1 decade ago
    Favorite Answer

    Since there's no phase change on reflection, the minimum path difference required between the direct and reflected waves should be λ, i.e. one time of wavelength itself.

    Now, λ = 340/200 = 1.7 m

    Also, the shortest reflection path is from the speaker to the mid-point on the wall corresponding to the line joining the microphone and the speaker.

    Then the length of the reflection path is:

    2 x square root of (x2 + 42) where x is the common distance of the microphone and speaker from the wall.

    Therefore we have, the path difference:

    2 x square root of (x2 + 42) - 8 = 1.7

    x2 + 16 = 23.5225

    x = 2.74 m

    Source(s): My physics knowledge
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