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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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There is friction between M and ground and $\;(\mu_{s}=0.5)\;$ . There is no friction between m and ground . Find the maximum amplitude of S.H.M with which the block m vibrates so that the block M does not move ? $\;(g=10 m/s^2)$


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Answer : (a) $\;33.3$
Explanation :
Let the amplitude of S.H.M be A
The spring streches maximum by 2A
$2KA=f \quad $ (f friction force )
$f=0.5 \times 40 \times 10$
$2 \times 3 \times A=0.5 \times 40 \times 10$
$A\times 2 \times 3 =5 \times 20 $
$A=\large\frac{100}{3}=33.3 \;.$
answered Mar 10, 2014 by yamini.v

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