Two particles A and B execute S.H.M.s of periods T and 2 5 T . If they start from the mean position, then the phase difference between them when the particle A completes two oscillations will be (in

Question

Two particles A and B execute S.H.M.s of periods T and 2 5 T ​ . If they start from the mean position, then the phase difference between them when the particle A completes two oscillations will be (in radian)

TestHub · Accepted Answer

Time taken by particle A to complete two oscillations is $t = 2 T$ .

The phase of a particle executing S.H.M. starting from the mean position at time $t$ is given by $ϕ = ω t = \frac{2 π}{T ^{'}} t$ , where $T^{'}$ is the time period.

For particle A, the phase at $t = 2 T$ is:
$ϕ_{A} = \frac{2 π}{T} \times 2 T = 4 π$

For particle B, the time period is $T_{B} = \frac{5 T}{2}$ . The phase at $t = 2 T$ is:
$ϕ_{B} = \frac{2 π}{\frac{5 T}{2}} \times 2 T = \frac{4 π}{5 T} \times 2 T = \frac{8 π}{5}$

The phase difference between the two particles is:
$Δ ϕ = ϕ_{A} - ϕ_{B} = 4 π - \frac{8 π}{5} = \frac{12 π}{5}$

Answer: $\frac{12 π}{5}$

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