Question

PLEASE HELP ASAP!!!

An oscillator with a frequency of 0.25 Hz is set in motion. One second later, an identical oscillator is set in motion. What is the phase separation of these two oscillators. Show your work please!

Answer 1

x1 = A sin ω t = A sin (2 pi f t) equation for SHM with no phase

x1 =A sin (2 * pi * .25 t) = A sin (.5 * pi * t)

x2 = A sin (.5 pi * (t - 1)) = A sin (.5 pi t - pi/2) second oscillator lags by 1 sec

So x2 is pi / 2 behind x1 in phase

Or x2 = A cos .5 pi t since sin (Ф - pi/2) = cos Ф

Answer 2

1Hz = One Cycle and One Cycle = 360degrees

When you want to find the difference caused by a time separation of two waves, you can find how much of a wave cycle will pass during the time delay.

At 0.25Hz, it takes 4 seconds to make 1 Cycle. With some algebra you find that 1 cycle every 4 seconds = a quarter cycle every 1 second.

So now that you know a one second delay causes a quarter of a cycle to pass, you can convert that into degrees by multiplying 0.25 and 360. 0.25x360=90

Lo and behold, a time difference of 1 second, creates a phase seperation of 90 degrees.