Two wavefunctions, given by y1(x,t) = 5.00sin[pi(4.00x-1200t)] y2(x,t) = 5.00sin

ID: 2243263 • Letter: T

Question

Two wavefunctions, given by

y1(x,t) = 5.00sin[pi(4.00x-1200t)]

y2(x,t) = 5.00sin[pi(4.00x-1200t-0.250)]

describe transverse waves traveling through some medium. Distances

are measured in meters.

(a) What wavefunction describes the resultant wave?

(b) What is the maximum amplitude of the resultant wave?

with the velocities of the individual waves?

(d) What is the transverse velocity of a piece of the medium, located at

position x and time t.

(e) What, then, is the transverse acceleration of a piece of the medium

at x=1.50 m and t=8.00 s?

(f) What part of this acceleration is due to the first wave alone? What

part is due to the second wave alone?

(g) If the phase difference between the two waves were zero

a) y = y1+y2=5(sin[pi(4.00x-1200t)]+sin[pi(4.00x-1200t-0.250)])

sin(a)+sin(b) = 2sin(a+b)/2cos(a-b)/2

sin[pi(4.00x-1200t)]+sin[pi(4.00x-1200t-0.250)]=2sin(pi(4x-1200t)-0.25*pi/2)cos(pi*0.25/2)

so y = 5*2sin(pi(4x-1200t)-0.25*pi/2)cos(pi*0.25/2)

==> y = 9.24sin(pi(4x-1200t)-0.25*pi/2)

b)Amax = 9.24

c) v = dy/dt = -9.24*1200cos(pi(4x-1200t)-0.25*pi/2) =

==> v = -11088cos(pi(4x-1200t)-0.25*pi/2)

d)transverse speed = dy/dt = -11088cos(pi(4x-1200t)-0.25*pi/2)

e)transverse acc= dv/dt = 11088*1200sin(pi(4x-1200t)-0.25*pi/2)

==> a = 11088*1200sin(pi(4*1.5-1200*8)-0.25*pi/2)=-5091832.67 units

f)a1 = 5*1200*1200sin(pi(4x-1200t))= 5*1200*1200sin(pi(4*1.5-1200*8))=1.86e-7

a2 = 5*1200*1200sin(pi(4x-1200t-0.25)=5*1200*1200sin(pi(4*1.5-1200*8-0.25)

==> a2=-5091168.82

the majority acc is due to wave 2

g)if phase differenmce =0

then y = 5*sin(pi(4x-1200t))+5*sin(pi(4x-1200t)) = 10*sin(pi(4x-1200t))

all the answers above would change is the phase difference is changed...

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