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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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The time taken by a particle in S.H.M to reach from equilibrium position to one end is 25 s . The angular frequency w is

$(a)\;8 \pi \times 10^{-2} \;s\qquad(b)\;2 \pi \times 10^{-2} \;s\qquad(c)\;6 \pi \times 10^{-2} \;s\qquad(d)\;4 \pi \times 10^{-2} \;s$

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1 Answer

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Answer : (b) $\;2 \pi \times 10^{-2} \;s$
Explanation :
Let time period be T
Time taken to reach from one extreme to equilibrium position is $\;\large\frac{T}{4}$
$\large\frac{T}{4}=$$25$
$T=100\;s$
$w=\large\frac{2\pi}{100}$$=2 \pi \times 10^{-2}\;.$
answered Mar 7, 2014 by yamini.v
edited Apr 2, 2014 by balaji.thirumalai
 

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