KCET · Physics · Oscillations
A particle executes SHM with amplitude \(0.2 \mathrm{~m}\) and time period \(24 \mathrm{~s}\). The time required it to move from the mean position to a point \(0.1 \mathrm{~m}\) from the mean position is
- A \(12 \mathrm{~s}\)
- B \(2 \mathrm{~s}\)
- C \(8 \mathrm{~s}\)
- D \(3 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
As the particle starts from the mean position,
\(\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}=\mathrm{A} \sin \frac{2 \pi}{\mathrm{T}} \mathrm{t}\)
or
\(0.1=0.2 \sin \frac{2 \pi}{24} t\)
or \(\quad \sin \frac{\pi}{12} \mathrm{t}=\frac{1}{2} \Rightarrow \sin \frac{\pi}{12} \mathrm{t}=\sin \frac{\pi}{6}\) or
\(\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}=\mathrm{A} \sin \frac{2 \pi}{\mathrm{T}} \mathrm{t}\)
or
\(0.1=0.2 \sin \frac{2 \pi}{24} t\)
or \(\quad \sin \frac{\pi}{12} \mathrm{t}=\frac{1}{2} \Rightarrow \sin \frac{\pi}{12} \mathrm{t}=\sin \frac{\pi}{6}\) or
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