First slide

Simple hormonic motion

Question

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

Moderate

A

zero
B

$\frac{π}{2}$
C

$\frac{π}{6}$
D

$\frac{π}{3}$

Solution

$x_{1} = a \sin (ωt + \emptyset_{1})$
$x_{2} = a \sin (ωt + \emptyset 2_{})$
$\Rightarrow | x_{1} - x_{2} | = 2 a \sin (ωt + \frac{\emptyset_{1} + \emptyset 2_{}}{2}) \cos (\frac{\emptyset_{1} + \emptyset_{2}}{2})$
To maximize $| x_{1} - x_{2} | : \sin (ωt + \frac{\emptyset_{1} + \emptyset_{2}}{2}) = 1$
$\Rightarrow a \sqrt{2} = 2 a \times 1 \times \cos (\frac{\emptyset_{1} - \emptyset_{2}}{2})$
$\Rightarrow \frac{1}{\sqrt{2}} = \cos (\frac{\emptyset_{1} - \emptyset_{2}}{2})$
$\Rightarrow \frac{π}{4} = \frac{\emptyset_{1} - \emptyset_{2}}{2} \Rightarrow \emptyset_{1} - \emptyset_{2} = \frac{π}{2}$

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