JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Hard
- Numerical
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 ......
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
General equation for displacement is given by
\(x=A \sin (\omega t+\phi)\)
\(\text { at given time }\)
\(\Rightarrow \omega t+\phi=30^{\circ}\)
\(\Rightarrow x=40 \times \frac{\sqrt{3}}{2} \Rightarrow 20 \sqrt{3}\,cm\)
\(\Rightarrow A=40\,cm\)
\(\Rightarrow K . E=\frac{1}{2} k\left(A^2-x^2\right)=200\)
\(200=\frac{1}{2} k\left(\frac{1600-1200}{100 \times 100}\right)\)
\(400 \times 100 \times 100=k \times 400\)
\(k=10^4\)
\(x=4\)
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