Two particles A and B execute S.H.M.s of periods T and (5T)/(2). If...

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

The phase of a particle executing S.H.M. starting from the mean position at time $t$ is given by $\phi = \omega t = \dfrac{2\pi}{T'}t$, where $T'$ is the time period.

For particle A, the phase at $t = 2T$ is:
$\phi_A = \dfrac{2\pi}{T} \times 2T = 4\pi$

For particle B, the time period is $T_B = \dfrac{5T}{2}$. The phase at $t = 2T$ is:
$\phi_B = \dfrac{2\pi}{\dfrac{5T}{2}} \times 2T = \dfrac{4\pi}{5T} \times 2T = \dfrac{8\pi}{5}$

The phase difference between the two particles is:
$\Delta\phi = \phi_A - \phi_B = 4\pi - \dfrac{8\pi}{5} = \dfrac{12\pi}{5}$

Answer: $\dfrac{12\pi}{5}$