Question

Two sinusoidal waves given below are superposed y1= a sin (kx-wt+π/6) & y2= a sin(kx-wt-π/6) equation of resultant wave is

Answer 1

Answer: $y_{1} + y_{2} = a\sqrt3\ sin(kx-\omega t)$

Explanation:

Given that,

$y_{1} = a sin(kx-\omega t +\dfrac{\pi}{6})$

$y_{2} = a sin(kx-\omega t -\dfrac{\pi}{6})$

Suppose that,

$A = kx-\omega t +\dfrac{\pi}{6}$

$B = kx-\omega t -\dfrac{\pi}{6}$

Now,

$A+B = 2(kx-\omga t)$

$\dfrac{A+B}{2} = kx-\omega t$

$A-B = 2(\dfrac{\pi}{6})$

$\dfrac{A-B}{2} = \dfrac{\pi}{6}$

The resultant wave  is

$y_{1} +y_{2} = a sin(kx-\omega t + \dfrac{\pi}{6}) + a sin(kx-\omega t - \dfrac{\pi}{6})$

$y_{1}+y_{2} = a(sinA+sinB)$.....(I)

General equation is

$SinA+SinB = 2 sin(\dfrac{A+B}{2}) cos (\dfrac{A-B}{2})$

Put the value of Sin A+ Sin B in equation (I)

$y_{1} + y_{2} = a[2sin(\dfrac{A+B}{2}) cos(\dfrac{A-B}{2})]$

Now, put the value of $\dfrac{A+B}{2}$ and $\dfrac{A-B}{2}$

$y_{1} + y_{2} = a[2sin(kx-\omega t) cos(\dfrac{\pi}{6})]$

$y_{1} + y_{2} = a[2sin(kx-\omega t) \times\dfrac{\sqrt 3}{2}]$

$y_{1} + y_{2} = a\sqrt3\ sin(kx-\omega t)$

Hence, the resultant wave is $y_{1} + y_{2} = a\sqrt3\ sin(kx-\omega t)$.

Answer 2

"Given:

$y_1 = a sin (kx -\omega t + \frac {\pi}{6})$

$y_2 = a sin (kx -\omega t - \frac {\pi}{6})$

Take,

$A = kx-\omega t + \frac {\pi}{6}$

$B = kx-\omega t - \frac {\pi}{6}$

Now, $A+B = 2(kx -\omega t)$

     $\frac { (A+B) }{ 2 } = kx -\omega t$

$A-B = 2\left( \frac { \pi }{ 6 } \right)$

$\frac {(A-B) }{2}=\frac{ \pi }{ 6 }$

Hence the resultant wave is,

$y_1 + y_2 = a sin (kx -\omega t + \frac {\pi}{6} ) + a sin (kx -\omega t - \frac {\pi}{6})$

$y_1 + y_2 = a (sin A + sin B) .....(I)$

The general equation is

$SinA+sinB=2sin\left( \frac { (A+B) }{ 2 } \right) cos\left( \frac { (A-B) }{ 2 } \right)$

Put the value of Sin A + Sin B in equation (I)

$y_1 + y_2 = a[2sin\left( \frac { (A+B) }{ 2 } \right) cos\left( \frac { (A-B) }{ 2 } \right)]$

Now, put the value of  $(\frac { (A+B) }{ 2 } \right)) and (\frac { (A-B) }{ 2 } \right)))$

$y_1 + y_2 = a [2sin (kx - \omega t) cos (\frac {\pi}{6})]$

$y_1 + y_2 = a [2sin (kx - \omega t ) \times\frac {\sqrt { 3 }}{2} ]$

$y_1 + y_2 = a\sqrt3 sin (kx - \omega t )$

Hence, the resultant $\omega$wave is

$y_1 + y_2 = a\sqrt3 sin (kx - \omega t ).$"