JEE Mains · Physics · STD 11 - 13. oscillations
At a given point of time the value of displacement of a simple harmonic oscillator is given as \(y = A \cos \left(30^{\circ}\right)\). If amplitude is \(40\,cm\) and kinetic energy at that time is \(200\, J\), the value of force constant is \(1.0 \times 10^{ x }\,Nm ^{-1}\). The value of \(x\) is ......
- A \(3\)
- B \(2\)
- C \(4\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
General equation for displacement is given by \(x=A \sin (\omega t+\phi)\) \(\text { at given time }\) \(\Rightarrow \omega t+\phi=30^{\circ}\) \(\Rightarrow x=40 \times \frac{\sqrt{3}}{2} \Rightarrow 20 \sqrt{3}\,cm\) \(\Rightarrow A=40\,cm\)…
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