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A rod of length 'l' and mass m, pivoted at one end, is held by a spring at its mid point and a spring at far end, both pulling in opposite directions. The springs have spring constant K, and equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position

Answer» <html><body><p><br/></p>Solution :Let `theta` be the <a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> <a href="https://interviewquestions.tuteehub.com/tag/angular-11524" style="font-weight:bold;" target="_blank" title="Click to know more about ANGULAR">ANGULAR</a> displacement of the rod <br/>Let `x_1 and x_2` be the changes in the lengths of the springs . Coresponding restoring forces will <a href="https://interviewquestions.tuteehub.com/tag/act-1106" style="font-weight:bold;" target="_blank" title="Click to know more about ACT">ACT</a> as shown. <br/>Here `x_2 = L theta and x_1 =(L theta)/(2)` <br/>Net restoning torque on the rod will be<br/>`tau =(K x_2)L + (Kx_1) L/2 =-(KL^2 + K (L^2)/(4))theta` <br/>So, `(ML^2)/(3) <a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a> = K ((5L^2)/(4)) theta` <br/>`alpha = - ((15L^2)/(4mL^2)K)theta , alpha =- ((15K)/(4m)) theta` <br/>frequency `=1/(<a href="https://interviewquestions.tuteehub.com/tag/2pi-1838601" style="font-weight:bold;" target="_blank" title="Click to know more about 2PI">2PI</a>) sqrt((alpha)/(theta)) = 1/(2pi) sqrt((15K)/(4m))` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_C_C05_E04_067_S01.png" width="80%"/></body></html>


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