Simple harmonic motion

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Question

Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance A/2 from mean position. Find phase difference in the two SHMs. (in degrees)

Moderate
Solution

Figure shows that two respective particles P' and Q' in SHM along with their corresponding particles in circular motion. Let P moves in upward direction when crossing Q' at A/2 as shown in Fig. (a), at this instant phase of P is ${\mathrm{\varphi }}_{1}={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{6}$                  …(i)Similarly, as shown in Fig. (b) we can take particle Q' is moving in downward direction (opposite P) at A/2, this implies its circular motion particle is in second quadrant thus its phase angle is${\mathrm{\varphi }}_{2}=\mathrm{\pi }-{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=\mathrm{\pi }-\frac{\mathrm{\pi }}{6}=\frac{5\mathrm{\pi }}{6}$              …(ii)As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (i) and (ii) as$\mathrm{\Delta \varphi }={\mathrm{\varphi }}_{2}-{\mathrm{\varphi }}_{1}=\frac{5\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{6}=\frac{2\mathrm{\pi }}{3}\mathrm{rad}={120}^{\circ }$

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The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers ${30}^{O}$ in 0.1s. The perpendicular projection ‘P’ from ‘A’ on the diameter MN represents the simple harmonic motion of ‘P’. The restoration force per unit mass when P touches M will be: