Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A linear harmonic oscillator of force constant \[2\times {{10}^{6}}N/m\] and amplitude 0.01m has total mechanical energy of 160 J. its: (1) maximum potential energy is 100 J (2) maximum kinetic energy is 100 J (3) maximum potential energy is 160 J (4) maximum potential energy is zeroA) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 are correct
D) 1, 2 and 3 are correct
Correct Answer: B
Solution :
\[K=\frac{1}{2}k{{A}^{2}}=\frac{1}{2}\times 2\times {{10}^{6}}\times {{(0.01)}^{2}}=100\,J\] This is basically the energy of oscillation of the particle. K,U and E at mean position (\[x=0\]) and extreme position\[(x=\pm A)\]are shown in figure. \[K=100\text{ }J=\]maximum\[K=0J\] \[U=60J=\]minimum \[U=160J=\]maximum \[E=160=\]constant \[E=160J=\]constantYou need to login to perform this action.
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