question_answer

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 zero

A)  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        \[E=160J=\]constant