A small source of sound executes simple harmonic motion with an amplitude of 17 cm . A detector is positioned along the source's line of motion. The source emits a sound of frequency 800 Hz that travels at a speed of 340 m s ⁻¹ . Determine the time period of the source if the width of the frequency band observed by the detector is 8 Hz .

Step-by-step solution

The maximum velocity of the source executing simple harmonic motion is v_ max = A = A ( 2 T ) . The maximum frequency observed by the detector occurs when the source moves towards it with maximum velocity: f_ max = f₀ ( v v - v_ max ) The minimum frequency observed occurs when the source moves away from it with maximum velocity: f_ min = f₀ ( v v + v_ max ) The width of the frequency band is given by f = f_ max - f_ min : f = f₀ v ( 1 v - v_ max - 1 v + v_ max ) = f₀ v ( 2v_ max v^2 - v_ max ^2 ) Since v_ max is mu