Two particles A and B are performing SHM along x and y-axis respectively with equal amplitude and frequency of 2 cm and 1 Hz respectively. Equilibrium positions of the particles A and B are at the co-ordinates (3, 0) and (0, 4) respectively. At t = 0, B is at its equilibrium position and moving towards the origin, while A is nearest to the origin and moving away from the origin. If the maximum and minimum distances between A and B is s 1 and s 2 then find s 1 + s 2 (in cm).
Text Solution
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At t = 0 Particle 2 is at point B and moving towards origin so displacement
Y = 4 – A sin ꞷt
Y = 4 – 2 sin ꞷt
and displacement of particle 1 is
X = 3 – A cos ꞷt
X = 3 – 2 cos ꞷt
So distance between them = 
s 2 = 29 – (16 sin ꞷt + 12 cos ꞷt) = 29 – 4 (4 sinꞷt + 3 cosꞷt)
⇒ 29 – 20 (sinꞷt + 37º)
So 
= 49 ⇒ s max = 7cm = S 1
= 9 ⇒ s min = 3cm =S 2 ∴ S 1 +S 2 = 10cm
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