Two points p₁ and p₂ are performing SHM along the same line about the same mean position. Initially they are at their positive extreme positions. If the time period of each point is 12 s and the difference of their amplitudes is 12 cm, then find the minimum time after which the separation between the points becomes 6 cm. (Assume points do not collide with each other as they pass through each other.)
Physics · Mixed Chapter
JEE Mains

Question

Two points p₁ and p₂ are performing SHM along the same line about the same mean position. Initially they are at their positive extreme positions. If the time period of each point is 12 s and the difference of their amplitudes is 12 cm, then find the minimum time after which the separation between the points becomes 6 cm. (Assume points do not collide with each other as they pass through each other.)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Because both points have the same period and start from positive extremes, their phases agree: x₁=A₁cosωt and x₂=A₂cosωt. Step 2: Separation is |x₁−x₂|=|A₁−A₂| |cosωt|=12|cosωt| cm. Step 3: For the first time the separation is 6 cm, cosωt=1/2, so ωt=π/3. Step 4: With ω=2π/12 rad s⁻¹, t=(π/3)/(π/6)=2 s.