Oscillations Question 4
Question 4 - 2024 (29 Jan Shift 1)
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$, where $x=$
Show Answer
Answer: (9)
Solution:
Let total energy $=E=\frac{1}{2} KA^{2}$
$U=\frac{1}{2} K\left(\frac{A}{3}\right)^{2}=\frac{KA^{2}}{2 \times 9}=\frac{E}{9}$
$KE=E-\frac{E}{9}=\frac{8 E}{9}$
Ratio $\frac{\text { Total }}{KE}=\frac{E}{\frac{8 E}{9}}=\frac{9}{8}$
$x=9$
