Question
A particle is executing simple harmonic motion with amplitude of $0.1 \mathrm{~m}$. At a certain instant when its displacement is $0.02$, its acceleration is $0.5 \mathrm{~ms}^{-2}$. The maximum velocity of the particle is (in $\mathrm{ms}^{-1}$ ) [BVP Engg. 2005](a) $0.01$(b) $0.05$(c) $0.5$(d) $0.25$
Step 1
Step 1: The equation of motion for a simple harmonic oscillator is given by: \[a = -\omega^2 x\] where \(a\) is the acceleration, \(\omega\) is the angular frequency, and \(x\) is the displacement. Show more…
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A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 m/s2. The maximum velocity of the particle is (in m/s)
A particle is executing simple harmonic motion with an amplitude of $0.02$ metre and frequency $50 \mathrm{~Hz}$ The maximum acceleration of the particle is: (a) $100 \mathrm{~m} / \mathrm{s}^{2}$ (b) $100 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}$ (c) $100 \pi \mathrm{m} / \mathrm{s}^{2}$ (d) $200 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}$
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(2) executing simple harmonic motion with A particle amplitude $5 \mathrm{~cm}$ and a time period $0.2 \mathrm{~s}$. The an and acceleration of the particle when the velocity and acco $\begin{array}{ll}\text { displacement is } 5 \mathrm{~cm} \text { is } \\ \text { (a) } 0.5 \pi \mathrm{m} \mathrm{s}^{-1}, 0 \mathrm{~m} \mathrm{~s}^{-2} & \text { (b) } 0.5 \mathrm{~m} \mathrm{~s}^{-1},-5 \pi^{2} \mathrm{~m} \mathrm{~s}^{-2}\end{array}$ (c) $0 \mathrm{~m} \mathrm{~s}^{-1},-5 \pi^{2} \mathrm{~m} \mathrm{~s}^{-2}$ (d) $0.5 \pi \mathrm{m} \mathrm{s}^{-1},-0.5 \pi^{2} \mathrm{~m} \mathrm{~s}^{-2}$
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