What is the phase difference between two simple harmonic motions represented by x1=A sin (ω t+(π/6)) and x2=A cos (ω t) ?

Question

What is the phase difference between two simple harmonic motions represented by x1=A sin (ω t+(π/6)) and x2=A cos (ω t) ? (A) (π/6) (B) (π/3) (C) (π/2) (D) (2π/3)

Tardigrade Admin · Accepted Answer

Given, x1=A sin (ω t+(π/6)) x2=A cos (ω t) x2=A cos t(ω t+(π/2)) Phase difference Δ φ=φ2-φ1 Δ φ=(π/2)-(π/6) Δ φ=(3 π-π/6)=(π/3)

Q. What is the phase difference between two simple harmonic motions represented by $x_{1} = A sin (ω t + \frac{π}{6})$ and $x_{2} = A cos (ω t)?$

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{2}$

$\frac{2 π}{3}$

Given,
$x_{1} = A sin (ω t + \frac{π}{6})$
$x_{2} = A cos (ω t)$
$x_{2} = A cos t (ω t + \frac{π}{2})$
Phase difference
$Δ ϕ = ϕ_{2} - ϕ_{1}$
$Δ ϕ = \frac{π}{2} - \frac{π}{6}$
$Δ ϕ = \frac{3 π - π}{6} = \frac{π}{3}$

Q. What is the phase difference between two simple harmonic motions represented by x1​=Asin(ωt+6π​) and x2​=Acos(ωt)?

6π​

3π​

2π​

32π​