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JEE Mains · Physics · STD 11 - 13. oscillations

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

  1. A \(\frac{\pi}{6}\)
  2. B \(\frac{\pi}{3}\)
  3. C \(\frac{\pi}{4}\)
  4. D \(\frac{\pi}{2}\)
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Answer & Solution

Correct Answer

(A) \(\frac{\pi}{6}\)

Step-by-step Solution

Detailed explanation

\(X = A \sin (\omega t +\delta)\) \(V = A \omega \cos (\omega t +\delta)\) \(\frac{ A }{2}= A \sin (\omega t +\delta)\) \(\therefore V \text { is tve, } \delta \text { must be }\) \(\text { At } t =0\) \(\text { in } 1^{\text {st }} \text { quadrant or } 4^{\text {th }}\)…
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