JEE Mains · Physics · STD 11 - 13. oscillations
When a particle of mass \(m\) is attached to a vertical spring of spring constant \(k\) and released, its motion is described by \(y ( t )= y _{0} \sin ^{2} \omega t ,\) where \('y'\) is measured from the lower end of unstretched spring. Then \(\omega\) is
- A \(\sqrt{\frac{g}{y_{0}}}\)
- B \(\sqrt{\frac{g}{2 y_{0}}}\)
- C \(\frac{1}{2} \sqrt{\frac{g}{y_{0}}}\)
- D \(\sqrt{\frac{2 g}{y_{0}}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{g}{2 y_{0}}}\)
Step-by-step Solution
Detailed explanation
\(y = y _{0} \sin ^{2} \omega t\) \(y =\frac{ y _{0}}{2}(1-\cos 2 \omega t )\) \(y -\frac{ y _{0}}{2}=-\frac{ y _{0}}{2} cos 2 \omega t\) Amplitude : \(\frac{y_{0}}{2}\) \(\frac{y_{0}}{2}=\frac{m g}{K}\) \(2 \omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{2 g}{y_{0}}}\)…
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