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Speed of particle executing simple harmonic motion with amplitude α is half of the maximum speed. At that instant, displacement of the particle is1. α/22. \(\frac{{\sqrt 3 }}{2}\alpha \)3. \(\frac{{2\alpha }}{{\sqrt 3 }}\)4. \(3\sqrt 2 \alpha\) |
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Answer» Correct Answer - Option 2 : \(\frac{{\sqrt 3 }}{2}\alpha \) Equations of simple harmonic motion are as followed \(\begin{array}{l}\dot x = A\omega \cos \omega t\\x = A\sin \omega t\end{array}\) Maximum speed will be when cosωt is maximum i.e 1 \(\Rightarrow {\left( {\dot x} \right)_{max}} = A\omega\) When speed is half of maximum \(\begin{array}{l}\Rightarrow \frac{{A\omega }}{2} = A\omega \cos \omega t \Rightarrow \cos \omega t = \frac{1}{2}\\\Rightarrow \omega t = \frac{\pi }{3}\end{array}\) At \(\omega t = \frac{\pi }{3}\), displacement will be \(x = A\sin \omega t = \alpha \sin \left( {\frac{\pi }{3}} \right) = \frac{{\sqrt 3 }}{2}\alpha \) |
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