Question

Two particles P and Q describe SHM of the same amplitude a and frequency f along the same straight line. The maximum distance between the two particles is $a\sqrt 2 $. The initial phase difference between the particles is
(A) $\dfrac{\pi }{3}$
(B) $\dfrac{\pi }{2}$
(C) $\dfrac{\pi }{6}$
(D) zero

Answer 1

Hint: When two particles move with the same amplitude and same frequency, the distance between them is maximum when they are equidistant from the mean position and their phase difference remains constant.

Complete step by step solution:
When relative distance between two particles is maximum or minimum, we can say that the differential of relative distance with respect to time will be zero. This means that the relative velocity of the particles should be zero, which suggests that the velocity of the particles should be the same.

When two particles are executing SHM along the same line, with same amplitude and same frequency, their velocity will be equal when they are moving in the same direction with equal speed. This is possible when the particles are on opposite sides of the mean position and are equidistant from the mean position.

From the above two conditions we can decide that both particles will be at equal distance from the mean position when the distance between them is maximum. Since the maximum distance between them is given to be $a\sqrt 2 $ , their respective positions from the mean position will be \[ + \dfrac{{a\sqrt 2 }}{2}\]and \[ - \dfrac{{a\sqrt 2 }}{2}\].

Putting these values in the general equation of SHM : $y = a\operatorname{Sin} (\omega t + \varphi )$
Putting the value of y for 1st particle,
$ + \dfrac{{a\sqrt 2 }}{2} = a\operatorname{Sin} (\omega t + {\varphi _1})$,
$ + \dfrac{{\sqrt 2 }}{2} = \operatorname{Sin} (\omega t + {\varphi _1})$
$\omega t + {\varphi _1} = \dfrac{\pi }{4}$
I.e phase of the first particle is $\dfrac{\pi }{4}$
Putting the value of y for 2nd particle,
\[ - \dfrac{{a\sqrt 2 }}{2} = a\operatorname{Sin} (\omega t + {\varphi _2})\],
$ - \dfrac{{\sqrt 2 }}{2} = \operatorname{Sin} (\omega t + {\varphi _2})$
$\omega t + {\varphi _2} = - \dfrac{\pi }{4}$
I.e phase of the second particle is $ - \dfrac{\pi }{4}$

Hence the phase difference between the particles is $\dfrac{\pi }{2}$

Therefore, the correct answer to the question is option : B

Note: The distance between two particles executing SHM is maximum when they are equidistant from the mean position. Similarly, when the speed of two particles in question is the same and the particles are executing similar SHMs then also their distance from the mean position is the same. This information helps in solving questions of two SHMs quickly.