Two particles 'A' and 'B' execute SHMs of periods 'T' and ′ 2 3 T ′ . If they start from the mean position then the phase difference between them, when the particle &#0

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Two particles 'A' and 'B' execute SHMs of periods 'T' and ′ 2 3 T ​ ′ . If they start from the mean position then the phase difference between them, when the particle 'A' completes two oscillations will be

TestHub · Accepted Answer

The angular frequencies of particles A and B are $ω_{A} = \frac{2 π}{T}$ and $ω_{B} = \frac{2 π}{\frac{3 T}{2}} = \frac{4 π}{3 T}$ .

The phase of a particle starting from the mean position is given by $ϕ = ω t$ .

When particle A completes two oscillations, the time elapsed is $t = 2 T$ .

The phase of particle A at $t = 2 T$ is $ϕ_{A} = ω_{A} t = \frac{2 π}{T} \times 2 T = 4 π$ .

The phase of particle B at $t = 2 T$ is $ϕ_{B} = ω_{B} t = \frac{4 π}{3 T} \times 2 T = \frac{8 π}{3}$ .

The phase difference between the two particles is $Δ ϕ = ϕ_{A} - ϕ_{B} = 4 π - \frac{8 π}{3} = \frac{4 π}{3}$ .

Answer: $\frac{4 π}{3}$

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