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[Maximum number: 2]

A mass is attached to a vertical spring. The other end of the spring is attached to the driver of an oscillator.

The mass is performing very lightly damped harmonic oscillations. The frequency of the driver is higher than the natural frequency of the system. At one instant the driver is moving downwards.

The oscillator is switched off. The system has a Q factor of 22. The initial amplitude is 10 cm . Determine the amplitude after one complete period of oscillation.

ALTERNATIVE 1:
$\begin{aligned} & \lessdot Q=2 \pi \frac{A_{0}^{2}}{A_{0}^{2}-A_{1}^{2}} \| \Rightarrow \frac{A_{1}^{2}}{A_{0}^{2}}=1-\frac{2 \pi}{Q} & \frac{A_{1}}{A_{0}}=\lessdot \sqrt{1-\frac{2 \pi}{22}}=» A_{1}=8.5 \lessdot \mathrm{~cm} \| \end{aligned}$

ALTERNATIVE 2:

driver amplitude is constant

so mass amplitude is unchanged at 10 cm

IB Physics HL C 4 5 Damping and Resonance Topic Practice

Question 1