Question 333: Oscillations

Question

A simple pendulum of length $$I$$ has a bob of mass $$m$$. It executes SHM of small amplitude A. The maximum tension in the string is ($$g=$$ acceleration due to gravity)

Accepted Answer

Consider the figure shown below
The string possess maximum tension when bolls at mean position of oscillation i.e., at position $$R$$.
From geometry, $$O P=\sqrt{1^2-A^2}$$
Also, $$\quad R P=O R-O P=I-\sqrt{1^2-A^2}$$
The whole kinetic energy of bob at position $$R$$ is converted into its potential energy at position $$B$$.
$$\begin{aligned}
	herefore \quad \frac{1}{2} m v^2 & =m g\left(I-\sqrt{I^2-A^2}ight) \
v^2 & =2 g\left(I-\sqrt{I^2-A^2}ight)
\end{aligned}$$
$$\begin{aligned}
&	ext { Balancing forces at } R 	ext {, }\
&\begin{aligned}
T-m g & =\frac{m v^2}{I}=\frac{2 m g\left(I-\sqrt{l^2-A^2}ight)}{I} \
	herefore \quad T & =m g+2 m g\left(1-\sqrt{1-\frac{A^2}{l^2}}ight)
\end{aligned}
\end{aligned}$$
$$	ext { Using approximation, } \sqrt{1-x^2}=1-\frac{x^2}{2} 	ext { for } x<<1 	ext {, we }$$
$$\begin{aligned}
T & =m g+2 m g\left[1-\left(1-\frac{A^2}{2^2}ight)ight] \
& =m g+m g\left(\frac{A}{I}ight)^2 \
& =m g\left[1+\left(\frac{A}{I}ight)^2ight]
\end{aligned}$$

Answer

$$m g$$

Answer

$$m g\left(\frac{A^2}{I^2}+1\right)$$

Answer

2 mg

Answer

$$m g\left(\frac{A}{I}+1\right)$$

Oscillations

A simple pendulum of length $I has a bob of mass m. It executes SHM of small amplitude A. The maximum tension in the string is (g=$ acceleration due to gravity)

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