Oscillations Question 11
Question 11 - 01 February - Shift 1
The amplitude of a particle executing SHM is 3$cm$. The displacement at which its kinetic energy will be $25 %$ more than the potential energy is:___________ $cm$.
Show Answer
Answer: (2)
Solution:
Formula: Total Mechanical Energy
$ \begin{aligned} & K E=P E+\frac{P E}{4} \\ & K E=\frac{5}{4} P E \\ & \frac{1}{2} m \omega^{2}(A^{2}-x^{2})=\frac{5}{4} \times \frac{1}{2} m \omega^{2} x^{2} \\ & {[v=\omega \sqrt{A^{2}-x^{2}}]} \\ & A^{2}-x^{2}=\frac{5}{4} x^{2} \\ & \frac{9 x^{2}}{4}=A^{2} \\ & x=\frac{2}{3} A \\ & \therefore x=\frac{2}{3} \times 3 cm \\ & \quad x=2 cm \end{aligned} $
