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I. Multiple Choice Questions · Q2

Q.A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3 s. The time period is a) 15 s b) 6 s c) 12 s d) 9 s

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Step 1. Represent the SHM as x=Rsin⁡(ωt+ε)x=R\sin(\omega t+\varepsilon) on the reference circle, with phase θ=ωt+ε\theta=\omega t+\varepsilon. Since SHM speed v=Rωcos⁡θv=R\omega\cos\theta depends only on cos⁡θ\cos\theta, 'crossing A and B with the same velocity' after a direct transit A→\toB of 3 s means cos⁡θ1=cos⁡(θ1+3ω)\cos\theta_1=\cos(\theta_1+3\omega), which forces θ1+3ω=−θ1+2nπ\theta_1+3\omega=-\theta_1+2n\pi for the non-trivial case, i.e. θ1=nπ−1.5ω\theta_1=n\pi-1.5\omega.

Step 2. So the phase at B is θ2=θ1+3ω=nπ+1.5ω\theta_2=\theta_1+3\omega=n\pi+1.5\omega.

Step 3. 'Returns to B after another 3 s' means the particle is again at the same position B, but now moving the opposite way (it has turned around at the extreme in between), 3 s later. Same position with opposite-signed velocity corresponds to phase θ3=π−θ2(mod2π)\theta_3=\pi-\theta_2 \pmod{2\pi} (since sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta)=\sin\theta but cos⁡(π−θ)=−cos⁡θ\cos(\pi-\theta)=-\cos\theta). Since θ3=θ2+3ω\theta_3=\theta_2+3\omega, this gives 2θ2+3ω=π+2mπ2\theta_2+3\omega=\pi+2m\pi.

Step 4. Substituting θ2\theta_2 from Step 2: 2(nπ+1.5ω)+3ω=π+2mπ⇒6ω=π+2(m−n)π2(n\pi+1.5\omega)+3\omega=\pi+2m\pi \Rightarrow 6\omega=\pi+2(m-n)\pi. Taking the smallest positive solution (m=nm=n): 6ω=π⇒ω=π/6 rad s−16\omega=\pi \Rightarrow \omega=\pi/6\ \text{rad s}^{-1}.

Step 5. Time period T=2π/ω=2π/(π/6)=12T=2\pi/\omega=2\pi/(\pi/6)=12 s.

✓Final answer

(c) 12 s

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