A simple pendulum with length 100,cm and bob | vedclass

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(C) The maximum tension in a simple pendulum occurs at the mean position,given by $T_{max} = mg + \frac{mv^2}{l}$.Using the conservation of energy,the

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$99$

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(C) The maximum tension in a simple pendulum occurs at the mean position,given by $T_{max} = mg + \frac{mv^2}{l}$.Using the conservation of energy,the kinetic energy at the mean position equals the potential energy at the extreme position: $\frac{1}{2}mv^2 = mgl(1 - \cos 	heta_0)$,where $\sin 	heta_0 = \frac{A}{l} = \frac{10}{100} = 0.1$.Thus,$\frac{mv^2}{l} = 2mg(1 - \cos 	heta_0)$.Substituting this into the tension formula: $T_{max} = mg + 2mg(1 - \cos 	heta_0) = mg(3 - 2\cos 	heta_0)$.Since $\cos 	heta_0 = \sqrt{1 - \sin^2 	heta_0} = \sqrt{1 - (0.1)^2} = \sqrt{0.99} \approx 1 - \frac{0.01}{2} = 0.995$.$T_{max} = 0.25 	imes 9.8 	imes (3 - 2 	imes 0.995) = 2.45 	imes (3 - 1.99) = 2.45 	imes 1.01 = 2.4745$.Given $T_{max} = \frac{x}{40}$,we have $x = 40 	imes 2.4745 = 98.98 \approx 99$.

$A$ simple pendulum with length $100\,cm$ and bob of mass $250\,g$ is executing $S$.$H$.$M$. of amplitude $10\,cm$. The maximum tension in the string is found to be $\frac{x}{40}\,N$. The value of $x$ is $..........$.

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