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(D) At the mean position,the tension $T$ is given by $T = mg + \frac{mv^2}{l}$,where $l$ is the length of the string.Given that $T = 3mg$,we have $mg

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(D) At the mean position,the tension $T$ is given by $T = mg + \frac{mv^2}{l}$,where $l$ is the length of the string.Given that $T = 3mg$,we have $mg + \frac{mv^2}{l} = 3mg$,which simplifies to $\frac{mv^2}{l} = 2mg$,so $v^2 = 2gl$.Let $	heta$ be the maximum angular displacement from the vertical. By the principle of conservation of energy,the kinetic energy at the mean position must equal the potential energy at the maximum displacement: $\frac{1}{2}mv^2 = mgl(1 - \cos 	heta)$.Substituting $v^2 = 2gl$,we get $\frac{1}{2}m(2gl) = mgl(1 - \cos 	heta)$.This simplifies to $mgl = mgl(1 - \cos 	heta)$,which implies $1 = 1 - \cos 	heta$,so $\cos 	heta = 0$.Therefore,$	heta = 90^\circ$.

$A$ simple pendulum oscillates in a vertical plane. When it passes through the mean position,the tension in the string is $3$ times the weight of the pendulum bob. What is the maximum displacement of the pendulum string with respect to the vertical in degrees?

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