Two sinusoidal waves given below are superposed y1=A sin (k x-ω t+(π/6)), y2=A sin (k x-ω t-(π/6)) The equation of resultant wave is

Question

Two sinusoidal waves given below are superposed y1=A sin (k x-ω t+(π/6)), y2=A sin (k x-ω t-(π/6)) The equation of resultant wave is (A) y=(A/√3) sin (k x-ω t) (B) y=A √3 sin (k x-ω t) (C) y=A √3 sin (k x-ω t-(π/3)) (D) y=(A/√3) sin (k x-ω t-(π/3))

Tardigrade Admin · Accepted Answer

y prime=y1+y2 or y'=A sin (k x-ω t+(π/6))+A sin (k x-ω t-(π/6)) or y'=2 A sin (k x-ω t) ⋅ cos ((((π/6)+(π/6))/2)) or y'=2 A (√3/2) sin (k x-ω t) ∴ y'=A √3 sin (k x-ω t)

Q. Two sinusoidal waves given below are superposed
$y_{1} = A sin (k x - ω t + \frac{π}{6}), y_{2} = A sin (k x - ω t - \frac{π}{6})$
The equation of resultant wave is

$y = \frac{A}{3} sin (k x - ω t)$

$y = A 3 sin (k x - ω t)$

$y = A 3 sin (k x - ω t - \frac{π}{3})$

$y = \frac{A}{3} sin (k x - ω t - \frac{π}{3})$

$y^{'} = y_{1} + y_{2}$
or $y^{'} = A sin (k x - ω t + \frac{π}{6}) + A sin (k x - ω t - \frac{π}{6})$
or $y^{'} = 2 A sin (k x - ω t) \cdot cos (\frac{( \frac{π}{6} + \frac{π}{6} )}{2})$
or $y^{'} = 2 A \frac{3}{2} sin (k x - ω t)$
$∴ y^{'} = A 3 sin (k x - ω t)$

Q. Two sinusoidal waves given below are superposed y1​=Asin(kx−ωt+6π​),y2​=Asin(kx−ωt−6π​) The equation of resultant wave is

y=3​A​sin(kx−ωt)

y=A3​sin(kx−ωt)

y=A3​sin(kx−ωt−3π​)

y=3​A​sin(kx−ωt−3π​)