What is the phase difference between two simple harmonic motions represented by $x 1 = A sin (ω t + \frac{π}{6})$ $x1=A sin(omegat +pi/6)$ and $x 2 = A cos ω t$ $x2=A cos omegat$ A. $\frac{π}{6}$ $pi/6$ B. $\frac{π}{3}$ $pi/3$ C. $\frac{π}{2}$ $pi/2$ D. $\frac{2 π}{3}$ $(2pi)/3$ ?

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Answer 1 · 2018-03-18T07:25:35

$B$ $B$ >

$A cos ω t$ $A cos omegat$ can be written as $A sin (\frac{π}{2} + ω t)$ $A sin(pi/2 +omegat)$

So, $\partial ϕ = (\frac{π}{2} + ω t - ω t - \frac{π}{6}) = \frac{π}{3}$ $del phi =(pi/2 +omegat -omegat-pi/6) =pi/3$

What is the phase difference between two simple harmonic motions represented by x1=A sin(omegat +pi/6)x1=Asin(ωt+π6)x1=A sin(omegat +pi/6) and x2=A cos omegatx2=Acosωtx2=A cos omegat A. pi/6π6pi/6 B. pi/3π3pi/3 C. pi/2π2pi/2 D. (2pi)/32π3(2pi)/3 ?