ExamBro
ExamBro
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 ......

  1. A \(3\)
  2. B \(2\)
  3. C \(4\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

General equation for displacement is given by \(x=A \sin (\omega t+\phi)\) \(\text { at given time }\) \(\Rightarrow \omega t+\phi=30^{\circ}\) \(\Rightarrow x=40 \times \frac{\sqrt{3}}{2} \Rightarrow 20 \sqrt{3}\,cm\) \(\Rightarrow A=40\,cm\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app