Download Homework 6 (Unit 3) Solutions (100 points)

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.