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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations2 Marks
Question
Two particles move periodically in the same straight line, their amplitude and frequency are the same. At the time when their displacement is half the amplitude, they cross each other going in opposite directions, what is the time difference between them?
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Answer

Taking the general equation of simple harmonic motion,
$y=A \sin (\omega t+\phi)$
Where $(\omega t +\phi)$ is called the total phase of the particle.
When $y=\frac{1}{2} A$ then $\frac{1}{2} A= A \sin (\omega t +\phi)$
$\begin{aligned}\frac{1}{2} & =\sin (\omega t+\phi) \\\omega t+\phi & =30^{\circ} \text { or } 150^{\circ}\end{aligned}$
At the moment both particles are moving in opposite directions, the phase of one of them is $30^{\circ}.$ So, the phase of second will be $180^{\circ}-30^{\circ}=150^{\circ}$. If one is $150^{\circ}$ then the other is $180^{\circ}-150^{\circ}=30^{\circ}$.
Hence the time difference at that moment
$=150^{\circ}-30^{\circ}=120^{\circ}$

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