Two simple pendulum $Δ$ have time periods ‘T’ and $5 T/4$ . They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when the pendulum with higher time period completes one oscillation is

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30°

60 °

90 °

45 °

answer is D.

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Detailed Solution

T_{1} = T; T_{2} = \frac{5T}{4}

$t = \frac{5T}{4}$

$Δ ϕ = ϕ_{1} - ϕ_{2}$

$Δ ϕ = 2 π (\frac{1}{T_{1}} - \frac{1}{T_{2}}) t$

$= 2 π (\frac{1}{T_{1}} - \frac{1}{\frac{5T}{4}}) \times \frac{5T}{4}$

$∵ ϕ = ω t; ϕ_{1} = \frac{2 π}{T_{1}} t$

$ϕ_{2} = \frac{2 π}{T_{2}} t$

$\Rightarrow Δ ϕ = \frac{2 π}{T} [1 - \frac{4}{5}] \times \frac{5T}{4}$

$= 2 π (\frac{1}{5}) \times \frac{5}{4} = \frac{π}{2} rad$

$Δ ϕ = 90 °$

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