The minimum phase difference between two simple harmonic motions is
$x_{1} = \frac{1}{2} sin ω t + \frac{1}{2} cos ω t$
$x_{2} = sin ω t + cos ω t$ $[sin \frac{π}{4} = cos \frac{π}{4} = \frac{1}{2}]$

Question

The minimum phase difference between two simple harmonic motions is x 1 ​ = 2 ​ 1 ​ sin ω t + 2 ​ 1 ​ cos ω t x 2 ​ = sin ω t + cos ω t [ sin 4 π ​ = cos 4 π ​ = 2 ​ 1 ​ ]

TestHub · Accepted Answer

The given equations of simple harmonic motion are:

$x_{1} = \frac{1}{2} sin ω t + \frac{1}{2} cos ω t$

$x_{1} = sin ω t cos \frac{π}{4} + cos ω t sin \frac{π}{4} = sin (ω t + \frac{π}{4})$

$x_{2} = sin ω t + cos ω t$

$x_{2} = 2 (\frac{1}{2} sin ω t + \frac{1}{2} cos ω t) = 2 sin (ω t + \frac{π}{4})$

The phase of $x_{1}$ is $ϕ_{1} = \frac{π}{4}$ and the phase of $x_{2}$ is $ϕ_{2} = \frac{π}{4}$ .

Phase difference $Δ ϕ = ϕ_{1} - ϕ_{2} = \frac{π}{4} - \frac{π}{4} = 0$ .

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