A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in same direction, then: (a) amplitude of motion is 3a (b) time period of oscillations is 8τ (c) amplitude of motion is 4a (d) time period of oscillations is 6τ

Answer: (d). Starting from rest means starting at an extreme: distance covered in time t is A(1 - cos ω t). After τ: A(1 - cos ωτ) = a; after 2τ: A(1 - cos 2ωτ) = 3a With c = cos ωτ: (1 - cos 2ωτ)/(1 - c) = (2(1-c)(1+c))/(1-c) = 2(1+c) = 3 ⇒ c = 1/2 ωτ = π/3 ⇒ T = (2π)/(ω) = 6τ (and A = 2a) So the answer is (d) T = 6τ.

A particle moves with simple harmonic motion in a straight line. In… | JEE Main 2014

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Periodic Motion & SHM Basics

A particle moves with simple harmonic motion in a straight line. In first $\tau s,$ after starting from rest it travels a distance $a,$ and in next $\tau s$ it travels $2a,$ in same direction, then:

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Answer: (d)

Starting from rest means starting at an extreme: distance covered in time $t$ is $A(1 - \cos\omega t)$ .
After $\tau$ : $A(1 - \cos\omega\tau) = a$ ; after $2\tau$ : $A(1 - \cos 2\omega\tau) = 3a$
With $c = \cos\omega\tau$ : $\frac{1 - \cos2\omega\tau}{1 - c} = \frac{2(1-c)(1+c)}{1-c} = 2(1+c) = 3 \Rightarrow c = \frac{1}{2}$
$\omega\tau = \frac{\pi}{3} \Rightarrow T = \frac{2\pi}{\omega} = 6\tau$ (and $A = 2a$ )
So the answer is (d) $T = 6\tau$ .

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