INSTRUCTIONS: (1) Please read the Reference Document: Sinusoidal Functions (foun

Question

INSTRUCTIONS: (1) Please read the Reference Document: Sinusoidal Functions (found in Canvas, in the same folder as this homework) before completing this activity. The Reference document has althe formulas you wil need, with examples. (2) Keep a copy of your work. Turn in one copy at the startofdass. Keep the other for reference. PROBLEM 1. This is a plot of two snusoidal functions of the same period Tbut different amplitude A. We will call them Function F1 and Function #2 Look carefully and keep track of which is which. The functons are ouT of PHASE (or out the period of the snusoidal Calculate the angular speed ofthe (c) have marked a place where you can clearly see the peaks of both plots. Directly from the graphs, read: Angle [rad] The time at which Function 1 has its The time at which Function 2 has its (d) So which one peakod first Functon 1 o This is the function that LEADS (e) Calauale the lint diference between the peaks: lat It tal (f The time diferenon, at is a fracson of the period Calculate what fraction of the period is it IAtl the number of radians that correspond to st Express as a multiple of (h) Now calculate also the comesponding number of degrees Show the calculation here CONCLUSION In this plot Function LEADS Function by degrees. Lab 11 Homework Page

Explanation / Answer

a)

The time period is the time difference between two maximas or two minimas. See that the function 1 attains maxima at t=0 and then again at t=12. This gives T=12s

b)

Angular speed is given by 2/T = 2*3.14/12 = 0.52 rad/s

c)

it can be seen from the graph that t1 = 12s and t2=10.5s

d)

Hence function 2 peaks first.

e)

t1-t2 = 1.5s

f)

(t1-t2)/T = 0.125s

g)

In radians = (t1-t2)/T *2 = 0.25*

h)

In degrees = 0.785 * (180/) = 45

Function 2 leads function 1 by 45 degrees.

Since function 2 leads, the phasor for function 2 will be further clockwise at an angle of 45 degrees with respect to phasor 1

PROBLEM 2

We follow the same steps as before. We see that function 1 peaks before function 2. Function 1 peaks at t=1/3 and function 2 peaks at t=1.

So dt = t1-t2 = 2/3s

Time period, T= 4s

Phase difference = dt/T *2 = /3 rad = 60 degrees

Hence function 1 leads function 2 by 60 degrees.

The phasor for function 2 at t=1 will be horizontal (since its phase is /2 seen from the sinusoids figure). Now since 1 leads 2 by 60 degrees, the phasor for function 1 will be 60 degrees clockwise with respect to phasor 2.

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