Use your calculator for algebraic calculations only. Solutions based on trial an

Question

Use your calculator for algebraic calculations only. Solutions based on trial and error, inspection of numerical plots, etc., are not acceptable. Consider the sinusoid x (t) = A cos (ohm t + phi), Where A > 0, and phi Element of ( - pi ,pi] Time t is in seconds. It is known that, x (t) >= 2.4 for exactly 18% of each period It takes 0.123 seconds for the value of the sinusoid to drop from 2.4 to the next minimum (valley) The first zero of the sinusoid with t > 0, occurs at t = 0.040s Determine the amplitude A Determine the period and angular frequency ohm of x (t). Determine the initial phase phi of x (t) as a fraction of pi. (Two possible values here).

Explanation / Answer

Part A

The sinusoid takes values greater than +2.4 for exactly 18%of a period. Thus beginning at a peak (height A), the sinusoid willdrop to the value +2.4 after a time interval corresponding to 9% of aperiod, or equivalently (in terms of angles) 9*pi/50 radians.

Therefore A*cos(9*pi/50) = 2.4, i.e., A = 2.4/cos(9*pi/50) = 2.8425

Part B

It takes 0.123 seconds to drop from 2.4 to the next minimum (=-A).This time interval is equivalent to an angle equal to

pi - 9*pi/50 = 41*pi/50 rad

  or 41% of a period.The period thus equals T = 0.123/0.41 = 0.3 seconds.

And the angular frequency is given by W = 2*pi/0.3 = 20*pi/3 rad/sec.

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