Two traveling sinusoidal waves are described by the wave function Y_1 = (5.00 m)

Question

Two traveling sinusoidal waves are described by the wave function Y_1 = (5.00 m)sin(pi .00x -1 200t)] and y_2 = (5.00 m)sin (pi (4.00times 1200t - 0.250)] Where x, y_1, and y_2 are in meters and t is in seconds, (a) What is the amplitude of the resultant wave? (b) What is the frequency of the resultant wave? Two sinusoidal waves combining in a medium are described by the wave functions Y_1 = (3 0cm) sin pi (x + 0.60t) and Y_2 =(3.0cm)sin pi(x -0.60t) Where x is in centimeters and t is in seconds Determine the maximum transverse position of an element of the medium at (a) x=0.250cm, (b) x 0.500 cm, and (c) x=l 50 cm. (d) Find the three smallest values of x corresponding to antinodes. Two speakers are driven by a common oscillator at 800 Hz and face each other at a distance of I 25 m. Locate the points along a line joining the two speakers where relative minima would be expected (use v=343 m/s). At 0.0891 m, 0.303 m, 0 158 m, 0.732 m, 0 947 m, and I 16 m from one speaker.

Explanation / Answer

1.) according to the problem,

A linear combination (i.e., a sum) of sin functions with the same period but different phases is also a sin function that has the same period, but a different phase. In general:

A*sin(Z) + B*sin(Z+D) = C*sin(Z+E) where

C = sqrt(A^2 + B^2 + 2*A*B*cos(D))

and

E = arctan[(B*sin(D))/(A+B*cos(D))]

Here, we have two sin functions:

y1 = a*sin(b*x - c*t)
and
y2 = a*sin(b*x - c*t - d)

with a = 5.0m (amplitude of individual waves)
b = pi*4.0 m^-1
c = pi*1200 sec^-1 (angular frequency)
d = pi*0.250

These are of the form given above, with:

A = B = a
Z = pi*b*x - pi*c*t
and
D = pi*d

The amplitude of the resulting wave is then:

a) E = sqrt(2*a^2 + 2*a^2*cos(pi*d)
E = sqrt(2*(5.0 m)^2 + 2*((5.0 m)^2)*cos(0.250*pi)
E = 9.24 m

b) The frequency of the wave is the same as the input waves. The angular frequency is pi*1200 Hz, which is related to the frequency by:

frequency = (angular frequency)/(2*pi)

so ,frequency = 600 Hz

2) According to the problem,

y= 3.00 sin (x + 0.600t) = 3 {sin x*cos 0.600t + cosx*sin 0.600t}
y= 3.00 sin (x - 0.600t),= 3 {sin x*cos 0.600t - cosx*sin 0.600t}

y1 + y2 = 6 (sin x)*(cos 0.600t )

For the maximum value (cos 0.600t ) = 1

a)At x = 0.25

6 (sin x)* = 4.24 cm

b) At x = 0.5

6 (sin x)* = 6 cm

c) At x = 1.5

6 (sin x)* = - 6 cm

d) y1 + y2 = 6 (sin x)*(cos 0.600t)

At anti nodes cos 0.600t = 1

y1 + y2 = 6 (sin x)*

sin x is the maximum when

x = 1/2, 3/2 , 5/2

3) According to the problem,

Two antinodes are required for a full wavelength in standing waves which are produced by the facing speakers: dNtoN = /2 or =2dNtoN

v        = f      

v        = f     2dNtoN

343    = 800(2dNtoN)

dNtoN = 0.214 meters

If the speakers vibrate in phase, the point halfway between them is anantinode of pressure at a distance from either speaker of

1.25 m / 2 = 0.625m

Then there is

a node at      0.625 – ½0.214 = 0.518 m

a node at      0.518 – 0.214 = 0.303 m

a node at      0.303 – 0.214 = 0.0891 m

a node at      0.518 + 0.214 = 0.732 m

a node at      0.732 + 0.214 = 0.947 m

& a node @   0.947 + 0.214 = 1.16 m from either speaker.

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