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Find the natural frequency of the system as shown in the figure. The pulleys are massless and frictionless. (Take k=9pi^(2)N//m and m=4kg)

Answer» <html><body><p></p>Solution :To find frequency we have to first calculate the acceleration of the block as a function of displacement. In such type of <a href="https://interviewquestions.tuteehub.com/tag/complicated-2533514" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLICATED">COMPLICATED</a> systems we can find this by <a href="https://interviewquestions.tuteehub.com/tag/using-7379753" style="font-weight:bold;" target="_blank" title="Click to know more about USING">USING</a> the equation of energy conservation and then <a href="https://interviewquestions.tuteehub.com/tag/differentiating-953151" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENTIATING">DIFFERENTIATING</a> it. One important point is to be noted is that the spring is already stretched by some amount and beyond this <a href="https://interviewquestions.tuteehub.com/tag/extension-981025" style="font-weight:bold;" target="_blank" title="Click to know more about EXTENSION">EXTENSION</a> we can not take energy to be equal to `(1//2)kx^(2)` so we have to first find out the initial extension. <br/> `mg=T_(1)`, `T_(1)=2T_(2)` <br/> `T_(2)=2T_(3)`, `T_(3)kx_(0)` <br/> `rArr x_(0)=(mg)/(4k)` <br/> If suppose the block is further pulley down by `x`, spring stretches by `4x` <br/> `U`(Total energy of system) `=(1)/(2)k(x_(0)+4x)^(2)-mgx+(1)/(2)mv^(2)` <br/> `(dU)/(dt)=0` <br/> `k(x_(0)+4x)4(dx)/(dt)-mg"(dx)/(dt)+mv(dv)/(dt)=0` <br/> `(dv)/(dt)=(d^(2)x)/(dt^(2))=-16(k)/(m)x` <br/> `rArr f=(1)/(2pi)sqrt((16k)/(m))rArrf=3`. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FIT_JEE_PHY_GMP_MOD2_C09_S01_047_S01.png" width="80%"/></body></html>


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