Question

A block undergoes simple harmonic motion about its equilibrium position ($x=0$) with amplitude $A$. Calculate fraction of the total energy is in the form of kinetic energy when the block is at position $x=\frac{1}{2}A$.

• A. $\frac{1}{3}$
• B. $\frac{3}{8}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$
• E. $\frac{3}{4}$

Answer 1

The correct option is E $\frac{3}{4}$

At the endpoints of the oscillation, the displacement (x) is equal to amplitude A. Since x is maximum so kinetic energy is zero and the total energy is in the form potential energy of spring i.e $\frac{1}{2}k{A}^2$.

It is also the expression of total energy at any other position. So, at $x=\frac{A}{2}$ we can write $KE+PE=E=\frac{1}{2}k{A}^2$

or $KE+\frac{1}{2}k(\frac{A}{2}{)}^2=\frac{1}{2}k{A}^2$ or $KE=\left(\frac{3}{8}\right)K{A}^2$

Thus, $\frac{KE}{E}=\frac{\left(\frac{3}{8}\right)k{A}^2}{\left(\frac{1}{2}\right)k{A}^2}=\frac{3}{4}$