Question 195: Work, Energy and Power & Circular Motion

Question

A simple pendulum oscillates with an angular amplitude $	heta$. If the maximum tension in the string is 4 times the minimum tension then the value of $	heta$ is

Accepted Answer

In simple pendulum,
$$ \mathrm{T}=\mathrm{mg} \cos \phi+\frac{\mathrm{mv}^2}{l} $$
At lowest point (Tension is maximum):
$$ \mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mv}^2}{l} \quad \ldots .\left(\phi=0^{\circ}ight) $$
At highest point (Tension is minimum):
$$ \mathrm{T}_{\min }=\mathrm{mg} \cos 	heta $$
$\ldots .($ angular amplitude $=	heta$ and $\mathrm{v}=0)$
By energy conservation,
$$ \begin{array}{ll}  & \frac{1}{2} \mathrm{mv}^2=\mathrm{mgl}(1-\cos 	heta) \ 	herefore & \frac{\mathrm{mv}^2}{l}=2 \mathrm{mg}(1-\cos 	heta) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\ & 	ext { But, } \mathrm{T}_{\max }=4 \mathrm{~T}_{\min } \ 	herefore & \mathrm{mg}+\frac{\mathrm{mv}^2}{l}=4 \mathrm{mg} \cos 	heta \ 	herefore & \mathrm{mg}+2 \mathrm{mg}(1-\cos 	heta)=4 \mathrm{mg} \cos 	heta \ 	herefore & 1+2-2 \cos 	heta=4 \cos 	heta \ 	herefore & 3=6 \cos 	heta \ \Rightarrow & \cos 	heta=\frac{1}{2} \ 	herefore & 	heta=\cos ^{-1}(0.5) \end{array} $$

Answer

$\cos ^{-1}(0.75)$

Answer

$\cos ^{-1}(0.5)$

Answer

$\sin ^{-1}(0.5)$

Answer

$\sin ^{-1}(0.75)$

Work, Energy and Power & Circular Motion

A simple pendulum oscillates with an angular amplitude \theta. If the maximum tension in the string is 4 times the minimum tension then the value of \theta is

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