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A loudspeaker produces a musical sound by the oscillation of a diaphragm. If the amplitude of oscillation is limited to 1.20 × 10 -3 mm, what frequencies will result in the acceleration of the diaphragm exceeding g?
Here the value of maximum acceleration ax of the diaphragm is equal to the acceleration due to gravity g and which has value 9.81 m/s2 on the surface of Earth.To obtain the frequency of diaphragm, substitute 9.81 m/s2 for acceleration ax, 1.20×10-3 mm for the amplitude of oscillation in the equation f = (√ ax/ xm) /2π,f = (√ ax/ xm) /2π = (√(9.81 m/s2 /1.20×10-3 mm) (1 mm/10-3 m )) /2×3.14=455/s= (455/s) (1 Hz/(1/sec))= 455 HzFrom the above observation we conclude that the frequency of diaphragm would be 455 Hz.
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