JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Medium
- MCQ
A particle executes \(S.H.M.\) of amplitude A along \(x\)-axis. At \(t =0\), the position of the particle is \(x=\frac{A}{2}\) and it moves along positive \(x\)-axis the displacement of particle in time \(t\) is \(x=A \sin (\omega t+\delta)\), then the value \(\delta\) will be
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
\(X = A \sin (\omega t +\delta)\) \(V = A \omega \cos (\omega t +\delta)\)
\(\frac{ A }{2}= A \sin (\omega t +\delta)\) \(\therefore V \text { is tve, } \delta \text { must be }\)
\(\text { At } t =0\) \(\text { in } 1^{\text {st }} \text { quadrant or } 4^{\text {th }}\)
\(\sin \delta=\frac{1}{2} \Rightarrow \delta=\frac{\pi}{6}, \frac{5 \pi}{6}\) \(\text { quadrant }\)
\(\therefore \text { Common solution is } \delta=\frac{\pi}{6}\)
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