Two transverse sinusoidal waves combining in a medium are described by the wave

Question

Two transverse sinusoidal waves combining in a medium are described by the wave functions y1 = 5.00 sin (x + 0.700t) y2 = 5.00 sin (x 0.700t) where x, y1, and y2 are in centimeters and t is in seconds. Determine the maximum transverse position of an element of the medium at the following positions.

(a) x = 0.130 cm
|ymax| =  cm

(b) x = 0.560 cm
|ymax| =  cm

(c) x = 1.30 cm
|ymax| =  cm

(d) Find the three smallest values of x corresponding to antinodes. (Enter your answers from smallest to largest.)

Explanation / Answer

here,

y1 = 5.00 sin (x + 0.700t)
y1 = 5 (sin x*cos 0.700t + cosx*sin 0.700t)

y2 = 5.00 sin (x 0.700t)
y2 = 5 {sin x*cos 0.700t - cosx*sin 0.700t}

y = y1 + y2
y = 10 (sin x)*(cos(0.600t) ) --------------------(1)

Part A:
x = 0.130 cm

for eqn 1 to be max , cos(0.600t) = 1,
From eqn 1,
ymax = 10*sin( *0.130)
ymax = 10*0.397
ymax = 3.97 cm

Part B:
x = 0.560 cm

for eqn 1 to be max , cos(0.600t) = 1,
From eqn 1,
ymax = 10*sin( *0.560)
ymax = 10*0.982
ymax = 9.82 cm

Part C:
x = 1.30 cm

for eqn 1 to be max , cos(0.600t) = -1,
From eqn 1,
ymax = 10*sin( *1.30)
ymax = -10 * 0.809
Ymax = - 8.09 cm

Part D:
the antidote occur when
x = n*lamda/4 , n = 1,2,3,4......

for combined wave function we find :
k = 2pi/lamda or lamda = 2

the smallest value of x corresponding to antidote are given by n = 1 , 3 , 5 ..

x1 = lamda/4 = 0.5 cm
x2 = 3*lamda/4 = 1.50 cm
x3 = 5*lamda/4 = 2.50 cm

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