The amplitude of the vibrating particle due to super position of two simple harmonic motions of y1= sin (ω t+(π/3)) and y2= sin (ω t) will be :

Question

The amplitude of the vibrating particle due to super position of two simple harmonic motions of y1= sin (ω t+(π/3)) and y2= sin (ω t) will be : (A) 2 (B)  √3  (C)  √2  (D) 1

Tardigrade Admin · Accepted Answer

For a wave travelling with amplitude a and angular velocity ω at time t, the displacement y is given by y=a sin ω t For phase difference φ, the general equation is y=a sin (ω t+φ) Given equations are y1= sin (ω t+(π/3)) and y2= sin ω t Comparing with standard equation, we have a1=1, a2=1, φ=(π/3)=60° Therefore, resultant amplitude is a=√a12+a22+2 a1 a2 cos φ a=√1+1+2 × 0.5=√3

Q. The amplitude of the vibrating particle due to super position of two simple harmonic motions of $y_{1} = sin (ω t + \frac{π}{3})$ and $y_{2} = sin (ω t)$ will be :

$2$

$3$

$2$

$1$

Q. The amplitude of the vibrating particle due to super position of two simple harmonic motions of y1​=sin(ωt+3π​) and y2​=sin(ωt) will be :

2

3​

2​

1

Q. The amplitude of the vibrating particle due to super position of two simple harmonic motions of $y_{1} = sin (ω t + \frac{π}{3})$ and $y_{2} = sin (ω t)$ will be :

$2$

$3$

$2$

$1$