Oscillations

JEE Main level, numerical answer, medium

78 of 194

Question 78 of 194

At a given point of time the value of displacement of a simple harmonic oscillator is given as y=Acos⁡(30∘).y=A\cos \left(30^{\circ}\right). If amplitude is 40cm40cm and kinetic energy at that time is 200J,200J, the value of force constant 1.0×10xNm–11.0\times {10}^{x}N{m}^{–1}. The value of xx is _____.

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Answer: 4

The formula to calculate the kinetic energy (K)\left(K\right) of a particle executing SHM is given by
K=12k(A2−y2)...(1)K=\dfrac{1}{2}k\left({A}^{2}-{y}^{2}\right)...\left(1\right)
Substitute the values of the known parameters into equation (1) and solve to calculate the value of the unknown parameter.
K=12×k×(A2−A2cos⁡230∘)=12×k×A2sin⁡230∘⇒200=12×1.0×10x×(0.4)2×(12)2=2×10x100⇒102=10x−2⇒x−2=2⇒x=4\begin{matrix}K & = & \dfrac{1}{2}\times k\times \left({A}^{2}-{A}^{2}{\cos }^{2}30^{\circ}\right) \\ & = & \dfrac{1}{2}\times k\times {A}^{2}{\sin }^{2}30^{\circ} \\ & \Rightarrow & 200=\dfrac{1}{2}\times 1.0\times {10}^{x}\times {\left(0.4\right)}^{2}\times {\left(\dfrac{1}{2}\right)}^{2} \\ & = & \dfrac{2\times {10}^{x}}{100} \\ & \Rightarrow & {10}^{2}={10}^{x-2} \\ & \Rightarrow & x-2=2 \\ & \Rightarrow & x=4\end{matrix}

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