ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
  • Hard
  • Numerical

A simple harmonic oscillator has an amplitude \(A\) and time period \(6 \pi\) second. Assuming the oscillation starts from its mean position, the time required by it to travel from \(x=A\) to \(x=\frac{\sqrt{3}}{2} A\) will be \(\frac{\pi}{x}\) s, where \(x=\) _______.

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

From phasor diagram particle has to move from \(\mathrm{P}\) to \(Q\) in a circle of radius equal to amplitude of \(SHM\)
\( \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \)
\( \phi=\frac{\pi}{6}\)
Now, \(\frac{\pi}{6}=\omega t\)
\(\frac{\pi}{6}=\frac{2 \pi}{T} t\)
\(\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t\)
\(\mathrm{t}=\frac{\pi}{2}\)
So, \(\mathrm{x}=2\)
Same subject
Explore more questions on app