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Home / Doubts and Solutions/JEE/Physics

Sir pls solve this question.

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Asked by rsudipto 14th December 2018, 9:49 AM
Answered by Expert
Answer:
Natural frequency of mass loaded spring system is begin mathsize 12px style fraction numerator 1 over denominator 2 straight pi end fraction square root of k over m end root end style , where k is force constant of spring and m is the loaded mass.
In the given system mass is loaded through pulleys. Hence instead of the force 'mg' acting on the spring, we have to consider the tension 
acting on the spring. Tension at various points are marked in the figure.
 
we have T1 = mg ;   2T2 = T1 or T2 = (1/2)mg  ;  2T3 = T2  or T3 = (1/4)mg 
 
Hence natural frequency = begin mathsize 12px style fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator k over denominator begin display style bevelled m over 4 end style end fraction end root space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator 4 k over denominator m end fraction end root end style
Answered by Expert 16th December 2018, 6:09 AM
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