ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2026
  • Medium
  • Numerical

The displacement of a particle, executing simple harmonic motion with time period \(T\), is expressed as \(x(t)= A \sin \omega t\), where \(A\) is the amplitude. The maximum value of potential energy of this oscillator is found at \(t=T / 2 \beta\). The value of \(\beta\) is ___________.

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

Potential energy is maximum at extreme position The particle starting at mean position reaches extreme position in time \(\frac{ T }{4}\).
So \(T/2\beta = T/4 \Rightarrow 2\beta = 4 \Rightarrow \beta = 2\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app