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A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross sectional area is 4.9 times 10^-7 m^2. IF the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad.s^-1. If the young's modulus of the material of the wire is n times 10^9 N.m^-2 Calculate the value of n.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/young-749721" style="font-weight:bold;" target="_blank" title="Click to know more about YOUNG">YOUNG</a> modules of the material of the <a href="https://interviewquestions.tuteehub.com/tag/wire-1457703" style="font-weight:bold;" target="_blank" title="Click to know more about WIRE">WIRE</a>, <br/> `Y=(F/A)/((DeltaA)/L)=(FL)/(A DeltaL) or,F=((YA)/L)DeltaL`......(1) <br/>IF mass m is pulled by a length `DeltaL` then restoring <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> developed in the wire, <br/> `F=k DeltaL`…….(2) <br/> Comparing equation (1) and (2) we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> , <br/> `k=(YA)/L`<br/>Angular frequency , `omega=sqrt(K/m)=sqrt((YA)/(<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>))`<br/> or,`140=sqrt((ntimes10^9times4.9times10^-7)/(0.1times1))=70sqrtn`<br/> `therefore sqrtn=2 or n=4`</body></html>


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