A simple pendulum of length l is made to osci | vedclass

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Question

(c)

For a simple pendulum, equation of tangential motion is

$F_T=-m g \sin 	heta$

$\Rightarrow m l \alpha=-m g \sin 	heta$

$\Rightarrow

Vedclass · Accepted Answer

slightly more than $T_0$

Vedclass · Answer

(c)

For a simple pendulum, equation of tangential motion is

$F_T=-m g \sin 	heta$

$\Rightarrow m l \alpha=-m g \sin 	heta$

$\Rightarrow \frac{d^2 	heta}{d t^2}+g \frac{l}{l} \sin 	heta=0$

Now, $\quad \sin 	heta=	heta-\frac{	heta^3}{3 !}+\ldots$

Approximately, $\sin 	heta \approx 	heta\left(1-\frac{	heta^2}{6}ight)$

Now, replacing $	heta_{ av }^2=\frac{1}{2} 	heta_0^2$

We have, $\frac{d^2 	heta}{d t^2}+\frac{g}{l}\left(1-\frac{	heta_0^2}{12}ight) 	heta=0$

$	herefore \quad \omega=\frac{g}{l}\left(1-\frac{	heta_0^2}{12}ight)$

So, time period of oscillation is

$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{g}{l}}\left[\left(1-\frac{	heta_0^2}{12}ight)^{-\frac{1}{2}}ight]$

Now, taking binomial approximately

$T \approx 2 \pi \sqrt{\frac{g}{l}} \cdot\left(1+\frac{	heta_l^2}{24}ight)$

Clearly, this is greater than $T_0=2 \pi \sqrt{\frac{g}{l}}$.

A simple pendulum of length $l$ is made to oscillate with an amplitude of $45$ degrees. The acceleration due to gravity is $g$. Let $T_0=2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be

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