AP EAMCET · PHYSICS · Oscillations
A point mass of \(400 \mathrm{~g}\) executes S.H.M. under a force \(\mathrm{F}=-\left(10 \mathrm{Nm}^{-1}\right) \mathrm{x}\). If it crosses the centre of oscillation with a speed of \(10 \mathrm{~ms}^{-1}\), the amplitude of motion is
- A \(2 \mathrm{~m}\)
- B \(4 \mathrm{~m}\)
- C \(0.4 \mathrm{~m}\)
- D \(0.5 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(4 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
We have, \(\mathrm{F}=-10 \mathrm{x}\) Comparing it with \(\mathrm{F}=-\mathrm{m} \omega^2 \mathrm{x}\), we get \(\mathrm{m} \omega^2=10\) \[ \Rightarrow \omega=\sqrt{\frac{10}{\mathrm{~m}}}=\sqrt{\frac{10}{0.4}}=5 \mathrm{rad} / \mathrm{sec} \] Now,…
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