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Question Number 45205 by Necxx last updated on 10/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

$$w=\sqrt{\frac{K}{m+\frac{I}{R^2}}}=\sqrt{\frac{K}{m+\frac{M}{2}}=\sqrt{\frac{1536}{5+1}}=\sqrt{256}=16}$$

Commented by Necxx last updated on 12/Oct/18

$$thanksforthehelp$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

$$tryingtoproveusingLagrangeeqn$$$$\frac{d}{dt}\left(\frac{\partial L}{\partial \stackrel{.}{q}}\right)-\frac{\partial L}{\partial q}=0$$$$L=T-V(T=k.EV=P.E)$$$$T=\frac{1}{2}m{\left(\stackrel{.}{x}\right)}^2+\frac{1}{2}\frac{I}{R^2}{\left(\stackrel{.}{x}\right)}^2$$$$\stackrel{.}{x}=\frac{dx}{dt}I=momentofinertiaofdisc$$$$w=\frac{\stackrel{.}{x}}{R}$$$$V=\frac{1}{2}{Kx}^{2}p.Eofspring$$$$L=T-V$$$$L=\frac{1}{2}m{\left(\stackrel{.}{x}\right)}^2+\frac{1}{2}\frac{I}{R^2}{\left(\stackrel{.}{x}\right)}^2-\frac{1}{2}{Kx}^2$$$$\frac{\partial L}{\partial \stackrel{.}{x}}=m\stackrel{.}{x}+\frac{I}{R^2}\stackrel{.}{x}$$$$\frac{d}{dt}\left(\frac{\partial L}{\partial \stackrel{.}{x}}\right)=(m+\frac{I}{R^2})\stackrel{..}{x}(\stackrel{..}{x}=\frac{d^{2}x}{{dt}^2})$$$$\frac{\partial L}{\partial x}=-Kx$$$$\frac{d}{dt}\left(\frac{\partial L}{\partial \stackrel{.}{x}}\right)-\frac{\partial L}{\partial x}=0$$$$(m+\frac{I}{R^2})\stackrel{..}{x}-kx=0$$$$\stackrel{..}{x}=\left(\frac{k}{m+\frac{I}{R^2}}\right)x$$$$w=\sqrt{\frac{k}{m+\frac{I}{R^2}}}$$$$\frac{2\pi }{T_1}=\sqrt{\frac{k}{m+\frac{I}{R^2}}}{T}_1=assumedtimeperiod$$$$T_1=2\pi .\sqrt{\frac{m+\frac{I}{R^2}}{K}}$$$$$$

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