An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the X axis. They are pulled from the equilibrium position in opposite directions along the X axis by a small angular amplitude 0= -1(0.9) and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is 330 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity g=10 m/s2) is [figure]
Answer
26 to 33
Solution
Maximum speed of each bob: v= 2gL(1- 0)= 2 10 8 0.1=4 m/s
Released from opposite sides together, they always move with equal and opposite velocities: at the mean position both approach (or both recede) at 4 m/s.
Approaching: f =660 330+4330-4=660 334326 676.2 Hz
Receding: f =660 326334 644.2 Hz
Maximum variation =f -f 32 Hz
Answer: 32