Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line, about the same mean position. The maximum distance between the two particles is a $\sqrt{2}$ . The phase difference between the particles is:

zero

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π / 2

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π / 6

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π / 3

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Solution

The correct option is B $π / 2$
$X_{1} = a s i n (ω t + ϕ_{1})$
$X_{2} = a s i n (ω t + ϕ_{2})$
$\Rightarrow | X_{1} - X_{2} | = 2 a c o s (ω t + \frac{ϕ_{1} + ϕ_{2}}{2})$ $s i n (\frac{ϕ_{1} - ϕ_{2}}{2})$
To maximize $| X_{1} - X_{2} | :$
$c o s (ω t + \frac{ϕ_{1} + ϕ_{2}}{2}) = 1$
$\Rightarrow a \sqrt{2} = 2 a \times 1 \times s i n (\frac{ϕ_{1} - ϕ_{2}}{2})$ $\Rightarrow \frac{1}{\sqrt{2}} = s i n (\frac{ϕ_{1} - ϕ_{2}}{2})$
$\Rightarrow \frac{π}{4} = \frac{ϕ_{1} - ϕ_{2}}{2}$ $\Rightarrow ϕ_{1} - ϕ_{2} = \frac{π}{2}$

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