JEE Mains · Physics · STD 11 - 13. oscillations
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 ___________.
- A 1
- B 2
- C 4
- D 8
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 \(T/2\beta = T/4 \Rightarrow 2\beta = 4 \Rightarrow \beta = 2\)
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