Question

{Use of Tech } Min​​imizing sound intensity Two sound speakers are 100 m apart and one speaker is three times as loud as the other speaker. At what point on a line segment between the speakers is the sound intensity the weakest? (Hint: Sound intensity is directly proportional to the sound level and inversely proportional to the square of the distance from the sound source.)

Accepted Answer

Define the sound intensity formula: Sound intensity (I) is proportional to the sound level (L) and inversely proportional to the square of the distance (d) from the source. Mathematically, this can be expressed as I = k * (L / $$d^2$$), where k is a proportionality constant. Let the sound level of the quieter speaker be L1 and the louder speaker be L2 = 3L1. Place the quieter speaker at position x = 0 and the louder speaker at x = 100 on a coordinate line. Consider a point P at a distance x from the quieter speaker and (100 - x) from the louder speaker. The sound intensity at point P due to the quieter speaker is I1 = k * (L1 / $$x^2) $$and due to the louder speaker is I2 = k * (3L1 / (100 - $$x)^2). $$The total sound intensity at point P is the sum of the intensities from both speakers: I_total = I1 + I2 = k * (L1 / $$x^2) + k * (3L1 / (100$$ - $$x)^2). To $$find the point where the sound intensity is weakest, differentiate I_total with respect to x, set the derivative equal to zero, and solve for x. This will give the critical points, which can be tested to find the minimum intensity.

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주요 개념

Sound Intensity

Inverse Square Law

Proportional Relationships