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Download Homework 6 (Unit 3) Solutions (100 points)
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Homework 6 (Unit 3) Solutions (100 points) 1. (15 points) The tip of a one-link robot is located at = 0 at time t=0 sec as shown in the figure. It takes 1 sec for the robot to move from = 0 to rad. If l =5 in, plot the x – and y- components as a function of time. Also find the amplitude, frequency, period, phase angle, and time shift. ’ SOLUTION: Amplitude = l = 5 in The period is equal to the time required to travel The frequency is determined as f = = = 1 Hz Also, ω = 2πf ω = 2π rad/s rad, thus T=1 sec The phase shift is equal to the initial position (in radians) Therefore, the time shift is also τ = 0 sec The equations of the tip are: x(t) = A cos (ωt + y(t) = A sin (ωt + = 5 cos (2πt) in = 5 sin (2πt) in 2. (15 points) The tip of a one-link robot is located at = rad at time t=0 sec as shown in the figure. The robot is rotating at an angular frequency of 2 If l =20cm, plot the x – and y- components as a function of time. Also find the amplitude, frequency, period, phase angle, and time shift. SOLUTION: Amplitude = l = 20 cm The frequency is determined from the angular frequency: ω = 2πf f= = = 1 Hz The period is then T= 1 sec The phase angle is equal to the initial position (in radians) Therefore, the time shift is also τ = τ = The equations of the tip are: x(t) = A cos (ωt + = 20 cos (2πt - ) cm y(t) = A sin (ωt + = 20 sin (2πt - ) cm = s (to the right) 3. (10 points) Use the trigonometry identity cos(A+B)=cosAcosB-sinAsinB to re-write this cosine function representing voltage. Mcos (120 + SOLUTION: Mcos (120 + = Mcos ( cos (120 + (-Msin( sin (120 4. (35 points) A sinusoidal current i(t) =0.1 sin (100t) amps is applied to the RC circuit shown in the figure below. The voltage across the resistor and capacitor are given by: vR (t) = 20 sin (100t) V vC (t) = -20 cos (100t) V where t is in seconds. a. The voltage applied to the circuit is given by v(t) = vR (t) + vC (t). Write v(t) in the form v(t) = M cos (100t + ) (i.e) find M and ) SOLUTION: v(t) = vR (t) + vC (t) = 20 sin (100t) -20 cos (100t) = M cos (100t + ) expanding the term on the RHS 20 sin (100t) -20 cos (100t) = M cos (100t ) cos ( ) – M sin (100t) sin ( ) now we compare coefficients on sin(100t) and cos (100t) on both sides, sin (100t) : 20 = -Msin Msin = -20 cos (100t) : -20 = Mcos Mcos = -20 M= √( ( = 180 + tan-1( ( ( = 28.28 V = 180 o + 45 o + 225o so v(t) = 28.28 cos (100t + 225 o) volts b. Suppose that v(t) = 50 cos (100t - ) volts. Write down the amplitude, frequency (in Hz),period (in seconds), phase angle (in degrees), and time shift (in seconds) of the voltage, v(t). SOLUTION: For v(t) = 50 cos (100t - ) Amplitude = 50 V Frequency = f f= = = 15.92 Hz = f The period is then T= 0.0628 sec The phase angle is equal to the initial position (in radians) Therefore, the time shift is τ = τ = = (to the right) c. Plot one cycle of the voltage v(t) = 50 cos (100t - ), and indicate the earliest time (after t=0) when the voltage is 50 volts. SOLUTION: 5. (25 points) A sinusoidal voltage v(t) = 10 sin (1000t) V is applied to the RLC circuit shown in the figure below. The current i(t) =0.707 sin (1000t + 45o) flowing through the circuit produces voltages across R, L, and C of vR (t) = 7.07 sin (1000t + 45o) V vL (t) = 7.07 sin (1000t + 135o) V vC (t) = 14.14 sin (1000t - 45o) V. a. Write down the amplitude, frequency (in Hz), period (in seconds), phase shift (in radians), and time shift (in m/sec) of the current i (t) = 7.07 sin (1000t + 45o) V. SOLUTION: Amplitude: A =0.707 A Frequency: ω = 2πf so f = Period: T= = = =159.15 Hz = 0.00628 sec Phase Shift: Φ = 45o Time Shift: τ= τ = = (to the left) b. Plot one cycle of the current i(t) =0.707 sin (1000t + 45o) A indicate the earliest time (after t=0) when the current is 0.707A.