Question #43057

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Question #43057

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Answer 1 · 2017-03-29T18:28:45

Assuming bigger pendulum as the pendulum of bigger time period i.e.having time period $\frac{5 T}{4}$ $(5T)/4$

As the two pendulums start at the same time in simple harmonic motion from the mean position,their initial phase angles are zero

The bigger pendulum completes its one oscillation in time $t = \frac{5 T}{4}$ $t=(5T)/4$ and comes back to the mean position.During this time the point on reference circle associated with SHM rotates by an angle $2 π$ $2pi$

If $ω$ $omega$ represents the angular velocity of the reference point associated with SHM of first pendulum of time period T,then the phase of the this pendulum after $\frac{5 T}{4}$ $(5T)/4$ s will be

$ω \frac{5 T}{4} = ω T + ω \frac{T}{4}$ $omega(5T)/4=omegaT+omegaT/4$

$= 2 π + \frac{2 π}{4} = 2 π + \frac{π}{2}$ $=2pi+(2pi)/4=2pi+pi/2$

Hence the phase difference

$2 π + \frac{π}{2} - 2 π = \frac{π}{2}$ $2pi+pi/2-2pi=pi/2$

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Question #43057

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