Oscillations

JEE Main level, numerical answer, medium

109 of 194

Question 109 of 194

A particle of mass 0.50 kg0.50 \mathrm{~kg} executes simple harmonic motion under force F=−50(Nm−1)xF=-50\left(\mathrm{Nm}^{-1}\right) x. The time period of oscillation is x35 s\frac{x}{35} \mathrm{~s}. The value of xx is _______
(Given π=227\pi=\frac{22}{7} )

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Answer: 22

m=0.5 kg F=−50(x)ma=(−50x)0.5a=−50xa=(−100x)W2=100⇒(w=10)T=2π10=(π5)=227×15=(2235)π35=2235⇒x=22\begin{aligned} & \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\end{aligned}

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