Question

Math Group 7 - Google Dr G Above what temper. Rot + 0/2 points Previous Answers SerPSE8 18.P.007.MI. Two traveling sinusoidal waves are described by the wave functions Y1 - 4.95 sin [(3.65x - 11350)] y2 = 4.95 sin [(3.65x - 11350 -0.250)] where X, Y. and Y2 are in meters and is in seconds. X (a) What is the amplitude of the resultant wave function yı + y2? 9.9 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all interm minimize roundoff error. m (b) What is the frequency of the resultant wave function? Hz Need Help? Readet Master It Talk to Tutor +-12 points SerPSE8 18.P.009. TWO sinusoidal waves in a string are defined by the wave functions Y1 = 1.80sin(18.0x - 31.01) Y2 = 1.80sin(28.0x - 43.00) where x, Y1, and y2 are in centimeters and is in seconds. (a) What is the phase difference between these two waves at the point x = 5.00 cm at t - 2.00 s? att-2.00 s? (At that location, the two waves add to Zero.) (b) What is the positive x value closest to the origin for which the two phases differ by t cm Need Help? Read It Talk to Tutor

Answer 1

By principal of superposition of two waves with same amplitude(1.8), frequency and wavelength with a phase difference ϕ(=pi/4) is given by ,

y ( x , t ) = y m sin ( kx - ωt ) + y m sin ( kx - ωt + ϕ )
= 2 y m cos ( ϕ / 2 ) sin ( kx - ωt + ϕ / 2 )

k = wave number
$\omega$ = angular frequency
2 y m cos ( ϕ / 2 ) is resultant amplitude

We can see by comparing,
(a) net amplitude = 2 * 4.95* Cos(pi/4 / 2) = 9.146

Angular frequency = $\omega$ = 1135 pi
2 pi frequency = 1135 pi
(b) frequency = 1135/2 = 567.5 hz

2nd part

phase is the term inside sin
(a)

phase difference = ( 28x - 43t )- (18x - 31t)
= 10x - 12t
= 10*5 - 12*2 = 50 - 24 = 26 (in radian )

(b)
+- pi = 10x-12(2) = 10x - 24
10x - 24 = +- 3.141
x = 2.714 cm , 2.085 cm
Closest to the origin is 2.085 cm