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 $\mathrm{x}=$
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Answer: (9)
Solution:
Let total energy $=\mathrm{E}=\frac{1}{2} \mathrm{KA}^{2}$
$\mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^{2}=\frac{\mathrm{KA}^{2}}{2 \times 9}=\frac{\mathrm{E}}{9}$
$\mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9}$
Ratio $\frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8}$
$\mathrm{x}=9$
