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Under the action of a force $F=-75 y$ where $F$ is in Newton and $y$ is in meters, an object of mass $3 \mathrm{~kg}$ executes simple harmonic motion. If the velocity of the object at the mean position is $2.5 \mathrm{~ms}^{-1}$, the maximum acceleration of the object is
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The correct answer is:
$12.5 \mathrm{~ms}^{-2}$
Action of force, $F=-75 y$
Compair with $\mathrm{F}=-\mathrm{k} \mathrm{y}$
$\mathrm{k}=75 \mathrm{~N} / \mathrm{m}$
Velocity of the object, $v=2.5 \mathrm{~m} / \mathrm{s}$
$\mathrm{V}_{\max }=\mathrm{A} \omega$
$\begin{aligned} & A=\frac{V_{\max }}{\omega}=V_{\max } \sqrt{\frac{m}{k}} \\ & =2.5 \times \sqrt{\frac{3}{75}}=0.5 \mathrm{~m}\end{aligned}$
Maximum acceleration, $a_{\max }=\omega^2 \mathrm{~A}$
$=\frac{\mathrm{k}}{\mathrm{m}} \times 0.5=\frac{75}{3} \times 0.5=12.5 \mathrm{~m} / \mathrm{s}^2$
Compair with $\mathrm{F}=-\mathrm{k} \mathrm{y}$
$\mathrm{k}=75 \mathrm{~N} / \mathrm{m}$
Velocity of the object, $v=2.5 \mathrm{~m} / \mathrm{s}$
$\mathrm{V}_{\max }=\mathrm{A} \omega$
$\begin{aligned} & A=\frac{V_{\max }}{\omega}=V_{\max } \sqrt{\frac{m}{k}} \\ & =2.5 \times \sqrt{\frac{3}{75}}=0.5 \mathrm{~m}\end{aligned}$
Maximum acceleration, $a_{\max }=\omega^2 \mathrm{~A}$
$=\frac{\mathrm{k}}{\mathrm{m}} \times 0.5=\frac{75}{3} \times 0.5=12.5 \mathrm{~m} / \mathrm{s}^2$
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