Question

The total energy of a harmonic oscillator of mass $2\text{kg}$ is $9\text{J}$. If its potential energy at mean position is $5\text{J}$, its $\text{K.E.}$ at the mean position will be :

• A. $9\text{J}$
• B. $14\text{J}$
• C. $4\text{J}$
• D. $11\text{J}$

Answer 1

The correct option is C $4\text{J}$

Since the total energy of a harmonic oscillator remains conserved. So $(T.E.)=(T.E.{)}_{atmean}=9\text{J}$.
Now,
$(T.E.{)}_{atmean}=(P.E.{)}_{atmean}+(K.E.{)}_{atmean}$
$\Rightarrow 9=5+(K.E.{)}_{mean}$
$(K.E.{)}_{atmean}=4\text{J}$
Hence, $\left(c\right)$ is the correct answer.