Find the frequency of a voltage generator

AI Thread Summary
To find the angular velocity "w" of a voltage generator, the correct approach involves calculating the period of the sine wave, which is determined by the output being 0.25 times Vmax after 0.0137 seconds. The period is calculated as 4 times 0.0137 seconds, resulting in 0.0548 seconds. Using the formula w = 2π/T, the angular velocity is found to be approximately 115.47 radians/sec. The next time the output reaches 0.25 Vmax is 0.0548 seconds later, totaling 0.0695 seconds. This method clarifies the relationship between the output and the sine wave period.
JamesL
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What am i doing wron guys?

"The output of a generator is given by:

V = (Vmax)sin(wt) ----> where "w" is omega

If after .0137 seconds, the output is .25 times Vmax, what is the largest possible angular velocity "w" of the generator?

What is the next time (value of "t") for which the output is .25 times Vmax?"

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now this problem seemed pretty straight forward to me. in order to calculate angular velocity "w" we need to know the period of the sine wave, which would be from Vmax to the next Vmax peak. since after .0137 secs it is 1/4 of Vmax... i figured it would take 16 of this .0137 chunks to equal the period. and then w is just 2*pi/period. this is incorrec though.

what am i doing wrong? is there a different place i should be starting from on the sine wave?
 
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The correct way to solve this problem is to use the formula w = 2π/T, where T is the period of the sine wave. Since you know the output is .25Vmax after .0137 seconds, you can calculate the period of the sine wave by dividing .0137 seconds by .25, giving you a period of 0.0548 seconds. Then you can use the formula to calculate the angular velocity, w = 2π/0.0548 = 113.59 radians/sec. The next time for which the output is .25Vmax is 0.0548 seconds later, so t = 0.0137 + 0.0548 = 0.0695 seconds.
 


It looks like you are on the right track, but there is a slight mistake in your calculation. The period of a sine wave is the time it takes for one complete cycle, which means going from one Vmax to the next Vmax. In this case, the period is not 16 of the .0137 chunks, but rather 4 of them. This is because the output is only .25 Vmax, not 1 Vmax. So, the correct calculation for angular velocity would be w = 2*pi/period = 2*pi/4(.0137) = 115.47 rad/s.

As for the next time for which the output is .25 Vmax, we can use the same logic. The period is still 4 of the .0137 chunks, so the next time would be after 4(.0137) = 0.0548 seconds. This would be the time for the next Vmax peak.

I hope this helps clarify your understanding of the problem. Keep up the good work!
 
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