A particle is executing simple harmonic motion with an amplitude of $2 \mathrm{~m}$. The difference in the…

A particle is executing simple harmonic motion with an amplitude of $2 \mathrm{~m}$. The difference in the magnitudes of its maximum acceleration and maximum velocity is 4 . The time-period of its oscillation and its velocity when it is 1 $\mathrm{m}$ away from the mean position are respectively
  1. $2 \mathrm{~s}, 2 \sqrt{3} \mathrm{~ms}^{-1}$
  2. $\frac{7}{22} \mathrm{~s}, 4 \sqrt{3} \mathrm{~ms}^{-1}$
  3. $\frac{22}{7} \mathrm{~s}, 2 \sqrt{3} \mathrm{~ms}^{-1}$
  4. $\frac{44}{7} \mathrm{~s}, 4 \sqrt{3} \mathrm{~ms}^{-1}$

Solution

As per question, $\begin{aligned} & \left|\mathrm{A} \omega^2\right|-|\mathrm{A} \omega|=4 \quad[\therefore \mathrm{A}=2 \mathrm{~m}] \\ & 2 \omega^2-2 \omega-4=0 \\ & \Rightarrow \omega=2 \mathrm{rad} / \mathrm{s} \Rightarrow \frac{2 \pi}{\mathrm{T}}=2 \\ & \Rightarrow \mathrm{T}=\pi=\frac{22}{7} s \end{aligned}$ Velocity, $v=\omega \sqrt{A^2-y^2}=2 \sqrt{(2)^2-(1)^2}$ $=2 \sqrt{4-1}=2 \sqrt{3} \mathrm{~ms}^{-1}$

Asked in: AP EAMCET 2016

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