Simple Harmonic Motion 1 Question 6

6. A linear harmonic oscillator of force constant $2 \times 10^{6} N / m$ and amplitude $0.01 m$ has a total mechanical energy of $160 J$. Its

$(1989,2 M)$

(a) maximum potential energy is $100 J$

(b) maximum kinetic energy is $100 J$

(c) maximum potential energy is $160 J$

(d) maximum potential energy is zero

Fill in the Blank

Show Answer

Answer:

Correct Answer: 6. $(b, c)$

Solution:

  1. $\frac{1}{2} k A^{2}=\frac{1}{2} \times 2 \times 10^{6} \times\left(10^{-2}\right)^{2}=100 J$

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.

$$ \begin{array}{ll} \hline & \\ x=0 & \\ K=100 J=\text { Maximum } & K=0 J \\ U=60 J=\text { Minimum } & U=160 J=\text { Maximum } \\ E=160 J=\text { Constant } & E=160 J=\text { Constant } \end{array} $$

$\therefore$ Correct options are (b) and (c).



NCERT Chapter Video Solution

Dual Pane