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I. Multiple Choice Questions · Q14

Q.A particle executes simple harmonic motion and displacement yy at time t0t_0, 2t02t_0 and 3t03t_0 are A, B and C, respectively. Then the value of A+C2B\dfrac{A+C}{2B} is

(a) cos⁡ωt0\cos\omega t_0
(b) cos⁡2ωt0\cos 2\omega t_0
(c) cos⁡3ωt0\cos 3\omega t_0
(d) 1
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Step 1. Write the SHM as y=Rsin⁡(ωt+ε)y=R\sin(\omega t+\varepsilon). Then A=Rsin⁡(ωt0+ε)A=R\sin(\omega t_0+\varepsilon), B=Rsin⁡(2ωt0+ε)B=R\sin(2\omega t_0+\varepsilon), C=Rsin⁡(3ωt0+ε)C=R\sin(3\omega t_0+\varepsilon).

Step 2. Using the sum-to-product identity sin⁡X+sin⁡Y=2sin⁡ ⁣(X+Y2)cos⁡ ⁣(X−Y2)\sin X+\sin Y=2\sin\!\left(\dfrac{X+Y}{2}\right)\cos\!\left(\dfrac{X-Y}{2}\right) with X=ωt0+εX=\omega t_0+\varepsilon and Y=3ωt0+εY=3\omega t_0+\varepsilon: X+Y2=2ωt0+ε\dfrac{X+Y}{2}=2\omega t_0+\varepsilon and X−Y2=−ωt0\dfrac{X-Y}{2}=-\omega t_0. …

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