MHT CET · Physics · Oscillations
A particle executes a simple harmonic motion of time period \(T\). Find the time taken by the particle to go directly from its mean position to half the amplitude
- A \(T / 2\)
- B \(\mathrm{T} / 4\)
- C \(T / 8\)
- D \(T / 12\)
Answer & Solution
Correct Answer
(D) \(T / 12\)
Step-by-step Solution
Detailed explanation
\(y=a \sin \omega t=\frac{a \sin 2 \pi}{T} t\)
At the mean position \(y=\frac{a}{2}\)
\(
\begin{array}{l}
\frac{a}{2}=\frac{a \sin 2 \pi t}{T} \\
t=\frac{T}{12}
\end{array}
\)
At the mean position \(y=\frac{a}{2}\)
\(
\begin{array}{l}
\frac{a}{2}=\frac{a \sin 2 \pi t}{T} \\
t=\frac{T}{12}
\end{array}
\)
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