5) Two speakers are driven in phase by a common oscillator at880 Hz and face eac

Question

5) Two speakers are driven in phase by a common oscillator at880 Hz and face each other at a distanceof 1.21 m. Locate the points along a linejoining the two speakers where relative minima of sound pressureamplitude would be expected. (Use v = 343 m/s. Choose onespeaker as the origin and give your answers in order of increasingdistance from this speaker. Minima Distance from
speaker (m) 1st 2nd 3rd 4th 5th 6th Minima Distance from
speaker (m) 1st 2nd 3rd 4th 5th 6th Minima Distance from
speaker (m) 1st 2nd 3rd 4th 5th 6th

Explanation / Answer

we know    wavelength = speed /freq = 343 / 880 =   0.389773 meters . and we also know, for minimal sound,    pathdifference = (n-1/2) wavelength . The distance from one speaker is x   while thedistance from the other speaker is 1.21 - x  so .      path difference =  1.21 - x    -   x    = 1.21 - 2x    and thisequals    .           1.21 -2x = (n -1/2) * 0.389773 . Simplify:        x = 0.605   -   (n-1/2) *0.194886 . we can now choose   n = -2, -1, 0, 1, 2,3          to find theminima. When we do we get the values .                    n           x .    first       -2           1.092    second   -1           0.897   third        0           0.702    fourth      1           0.508   fifth         2           0.313   sixth        3           0.118

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