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I. Multiple Choice Questions · Q7

Q.The displacement of a simple harmonic motion is given by y(t)=Asin⁡(ωt+ϕ)y(t) = A \sin(\omega t + \phi) where A is amplitude of the oscillation, ω\omega is the angular frequency and ϕ\phi is the phase. Let the amplitude of the oscillation be 8 cm and the time period of the oscillation is 24 s. If the displacement at initial time (t=0t = 0 s) is 4 cm, then the displacement at t=6t = 6 s is

(a) 8 cm
(b) 4 cm
(c) 43\dfrac{4}{\sqrt3} cm
(d) 434\sqrt3 cm
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Step 1. ω=2π/T=2π/24=π/12 rad s−1\omega=2\pi/T=2\pi/24=\pi/12\ \text{rad s}^{-1}.

Step 2. At t=0t=0: y(0)=Asin⁡ϕ=4⇒8sin⁡ϕ=4⇒sin⁡ϕ=1/2⇒ϕ=π/6y(0)=A\sin\phi=4 \Rightarrow 8\sin\phi=4\Rightarrow\sin\phi=1/2\Rightarrow\phi=\pi/6 (30 degrees). …

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