1T=\pi \sqrt{\frac{I}{K_{2}a^{2}}}+\pi \sqrt{\frac{4I}{K_{1}L^{2}}}
6 Answers
23\omega =\sqrt\frac{3(k_{1}+k_{2})}{m}??
consiering that x=a=L/2
bcz upper length is L/2 in d fig ..
23dimensions itself are wrong @ joker
btw donno why did i take x=L/2 [16]
23T=\pi \sqrt{\frac{4I}{k_{1}L^{2}}}+\pi \sqrt{\frac{I}{k_{2}a^{2}}} ??
btw realised da mistake dat i made initially [4]
