A particle of mass $0.50 \mathrm{~kg}$ executes simple harmonic motion under force $\mathrm{F}=-50\left(\mathrm{Nm}^{-1}\right) \mathrm{x}$. The time period of oscillation is $\frac{x}{35} s$. The value of $x$ is . . . . .(Given $\pi=\frac{22}{7}$ )
JEE MAIN 2024, Diffcult
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$\mathrm{m}=0.5 \mathrm{~kg}$
$\mathrm{~F}=-50(\mathrm{x})$
$\mathrm{ma}=(-50 \mathrm{x})$
$0.5 \mathrm{a}=-50 \mathrm{x}$
$\mathrm{a}=(-100 \mathrm{x})$
$\mathrm{W}^2=100 \Rightarrow(\mathrm{w}=10)$
$\mathrm{T}=\frac{2 \pi}{10}=\left(\frac{\pi}{5}\right)=\frac{22}{7 \times 15}=\left(\frac{22}{35}\right)$
$\frac{\pi}{35}=\frac{22}{35} \Rightarrow \mathrm{x}=22$
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