A particle executing S.H.M starts from the mean position. Its amplitude is ' $\mathrm{A}$ ' and time period…

A particle executing S.H.M starts from the mean position. Its amplitude is ' $\mathrm{A}$ ' and time period ' $\mathrm{T}$ ' At what displacement its speed is one-fourth of the maximum speed?
  1. $\frac{\mathrm{A}}{\sqrt{15}}$
  2. $\frac{\mathrm{A}}{4}$
  3. $\frac{4 \mathrm{~A}}{15}$
  4. $\frac{\mathrm{A} \sqrt{15}}{4}$

Solution

$\begin{aligned} & \mathrm{V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & \mathrm{~V}_{\max }=\mathrm{A} \omega \\ & \therefore \sqrt{\mathrm{A}^2-\mathrm{x}^2}=\frac{\mathrm{A}}{4} \\ & \therefore 16 \mathrm{~A}^2-16 \mathrm{x}^2=\mathrm{A}^2 \\ & \therefore 16 \mathrm{x}^2=15 \mathrm{~A}^2 \\ & \therefore \mathrm{x}^2=\frac{15}{16} \mathrm{~A}^2 \\ & \therefore \mathrm{x}=\frac{\mathrm{A} \sqrt{15}}{4}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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