Two particles A and B execute simple harmonic motion of period T and 5 T / 4 . They start from mean position. The phase difference between them when the particle A completes one oscillation will be

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TestHub · Accepted Answer

Equation of motion of the particles are
$X_{1} = A_{1} \sin \frac{2 π}{T_{1}} t$
and $X_{2} = A_{2} \sin \frac{2 π}{T_{2}} t$
$∴$ Phase difference $Δ ϕ = (\frac{2 π}{T_{1}} - \frac{2 π}{T_{2}}) t$
$= (\frac{2 π}{T} - \frac{2 π}{5 T / 4}) t$
at $t = T$
$Δ ϕ = (2 π - \frac{4 \times 2 π}{5}) \frac{T}{T} = \frac{2 π}{5}$

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