Hello everyone! I have a problem simple harmonic motion. Below is the question:

Question

Hello everyone! I have a problem simple harmonic motion.

Below is the question:

These two waves travel along the same string:

What are (a) the amplitude and (b) the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.36 mm is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

Below is the hint:

You need to represent each wave by a phasor (a vector). Do you see that the phase angle gives the angle of a vector relative to the first vector? How should the third vector (the third phasor) be oriented relative to the sum of the first two in order to give the longest resultant vector?

And I have tried once, below is my previous answer:

Thanks for your help!

y1 = (3.68 mm) sin(2.23x - 490t) y2 = (5.46 mm) sin(2.23x - 490t + 0.783rad). (a) Number To.011 (b) NumberTT1.37 (c) Number TT5.37 UnitsT mm UnitsT rad Units TT mm

Explanation / Answer

y = y1 + y2

y = 3.68 sin(2.23x - 490t) + (5.56 mm) sin(2.23x - 490t + 0.783rad)

y = 3.68*sin(2.23x - 490t) + (5.56 mm) sin(2.23x - 490t )*cos0.783 + (5.56 mm) cos(2.23x - 490t )*sin0.783

y = sin(2.23x - 490t)*[3.68 + (5.56 *cos0.783) ] + cos(2.23x - 490t )* (5.56*sin0.783 )

R*cos = 3.68 + (5.56 *cos0.783)

R*sin = (5.56*sin0.783 )

y12 = R*sin(2.23x - 490t+)

part (a):

R = sqrt(3.682+5.562+(2*3.68*5.56*cos(0.783 ))
R = 3.5610

part (b):

phase angle = tan-1(R*sin / R*cos)

phase angle = tan-1((5.56*sin0.783 )/(3.68 + (5.56 *cos0.783)))

phase angle = -1.390 radian

part (c):

To maximize the amplitude of the new resultant wave, phase angle should be 1.390 radian

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.