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  • A particle is performing angular SHM has amplitude as  . If at <img alt="t=\frac{T}{4}" src="https://learn.care

A particle is performing angular SHM has amplitude as \pi . If at t=\frac{T}{4} its angular displacement was \frac{\sqrt{3}\pi }{2} then what is phase constant of SHM

Option: 1

\frac{\pi}{3}


Option: 2

\frac{\pi}{6}


Option: 3

\frac{\pi}{2}


Option: 4

\frac{\pi}{4}


Answers (1)

best_answer

 

 

 

 

  The general equation of angular SHM is given by \theta =\theta _0sin(\omega t+ \phi )

From the question we have

 \\\frac{\sqrt {3}\pi}{2} = \pi sin( \frac{2\pi}{T} *\frac{T}{4}+ \phi )\\ \\ \Rightarrow \frac{\sqrt {3}}{2}=sin(\frac{\pi}{2}+\phi)=Cos(\phi) \\ \\ \Rightarrow \phi=\frac{\pi}{6}

Posted by

vishal kumar

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