You and your friend run around a racetrack at constant speed but you are faster than your friend. While you complete a lap in time T1, your friend completes it in a longer time T2. Suppose that at time t=0 the two of you are abreast and that you are just about to overtake him. Suppose further that the next time you pass him he has run n full laps. Prove that your passing frequency f, i.e. the number of times per second that you pass him is given by (T2-T1)/(T1*T2).
2007-07-09 08:27:50 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You're essentially measuring beat frequency here. Your frequency is
f1 = 1/(T1)
while his is
f2 = 1/(T2).
The beat frequency (passing frequency in this case) is simply the difference in the two frequencies:
fb = f2 - f1
fb = 1/(T2) - 1/(T1)
fb = (T2-T1) / (T1*T2)
2007-07-09 08:38:27 · answer #1 · answered by lithiumdeuteride 7 · 0⤊ 0⤋
your frequency f1 = 1/T1
friend's frequency f2 = 1/T2
relative frequency = f1 - f2
= 1/T1 - 1/T2 = (T2-T1)/(T1 T2)
2007-07-09 15:34:00 · answer #2 · answered by B B 4 · 1⤊ 0⤋