JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2021
- Medium
- MCQ
A particle starts simple harmonic motion from the mean position. Its amplitude is \(a\) and total energy \(E\). At one instant its kinetic energy is \(3E/4.\) Its displacement at that instant is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(y=\frac{a}{2}\)
Step-by-step Solution
Detailed explanation
(b) \(\frac{K}{E} = \frac{{\frac{1}{2}m{\omega ^2}({a^2} - {y^2})}}{{\frac{1}{2}m{\omega ^2}{a^2}}}\)
\(= \frac{{{a^2} - {y^2}}}{{{a^2}}} \)
\(= 1 - \frac{{{y^2}}}{{{a^2}}}\)
So, \(\frac{{\left( {\frac{{3E}}{4}} \right)}}{E} = 1 - \frac{{{y^2}}}{{{a^2}}}\)
\(\Rightarrow \frac{{{y^2}}}{{{a^2}}} = 1 - \frac{3}{4} = \frac{1}{4}\)
\( \Rightarrow y = \frac{a}{2}\).
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