Answer: y'=2Asin(kx)cos(wt)
Explanation:
Let y1=A sin (kx + wt) be the first wave
y2=A sin (kx - wt) be the second wave in the opposite direction (which we showed by putting a negative sign between the terms kx and wt)
Please do note that both wave have the same attributes (that's Amplitude, wave number and angular frequency) because they are formed on the same medium by the same source just that their directions are opposite.
By super imposing these 2 waves, we have a resulting singular wave representing both wave (law of superimposition) with a resulting value of vertical displacement y'.
Thus y' = y1 + y2.
Let us do the math.
y'=A sin (kx + wt) + A sin (kx - wt)
By factoring A out, we have that
y' = A [ sin (kx + wt) + sin (kx - wt)]
For simplicity let us use the substitution
Let (kx + wt) = a and (kx - wt) =b
Hence we have that
y' = A [sin a + sin b].
From trigonometric ratio
sin a + sin b = 2sin[(a+b)/2] * cos [(a - b)/2]
By recalling that (kx + wt) = a and (kx - wt) =b
sin a + sin b = 2sin [(kx +wt +kx-wt) /2] * cos [(kx +wt - (kx-wt))/2]
Thus we have that
sin a + sin b = 2sin [(kx+wt+kx-wt)/2] * cos[(kx+wt-kx+wt)/2]
By collecting like terms in the bracket we have that
sin a + sin b = 2sin[2kx/2] * cos [2wt/2]
By dividing
sin a + sin b = 2sin(kx) cos(wt)
Now let us get the final resultant vertical displacement (y')
Recall that
y' = A [sin a + sin b]. and we already deduced that
sin a + sin b = 2sin(kx) cos(wt)
Finally,
y' = A [2sin(kx) cos(wt)] which is
y'=2Asin(kx)cos(wt)...... Final answer
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