Two particles P and Q describe simple harmoni | vedclass

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(D) Let the displacement of the particles be given by $x_P = A \sin(\omega t + \phi_1)$ and $x_Q = A \sin(\omega t + \phi_2).$ The velocity is $v = \o

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$2$

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(D) Let the displacement of the particles be given by $x_P = A \sin(\omega t + \phi_1)$ and $x_Q = A \sin(\omega t + \phi_2).$ The velocity is $v = \omega \sqrt{A^2 - x^2}.$Case $1$: When $P$ and $Q$ are on opposite sides at the same distance,$x_P = -x_Q = x.$ Their velocities are $v_P = \omega \sqrt{A^2 - x^2}$ and $v_Q = \omega \sqrt{A^2 - x^2}.$ Given $v_P = v_Q = 1.2 \, m/s,$ this implies $\omega \sqrt{A^2 - x^2} = 1.2.$ Using the phasor diagram,this corresponds to the vertical component of the velocity vector,$v = v_{max} \cos 	heta = 1.2.$Case $2$: When displacements are the same,$x_P = x_Q = x',$ their velocities are $v_P = \omega \sqrt{A^2 - (x')^2}$ and $v_Q = -\omega \sqrt{A^2 - (x')^2}$ (opposite directions). Given $|v_P| = |v_Q| = 1.6 \, m/s,$ this corresponds to $v = v_{max} \sin 	heta = 1.6.$Squaring and adding the two equations: $(v_{max} \cos 	heta)^2 + (v_{max} \sin 	heta)^2 = (1.2)^2 + (1.6)^2.$$v_{max}^2 = 1.44 + 2.56 = 4.0.$$v_{max} = \sqrt{4} = 2 \, m/s.$

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