Question 352: Oscillations

Question

A bob of a simple pendulum has mass $m$ and is oscillating with an amplitude $a$. If the length of the pendulum is $L$, then the maximum tension in the string is $\left[\cos 0^{\circ}=1\right.$, $g=$ acceleration due to gravity]

Accepted Answer

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

Answer

$m g\left[1+\left(\frac{a}{L}\right)^2\right]$

Answer

$m g\left[1-\left(\frac{L}{a}\right)^2\right]$

Answer

$m g\left[1+\left(\frac{L}{a}\right)^2\right]$

Answer

$m g\left[1-\left(\frac{a}{L}\right)^2\right]$

Oscillations

A bob of a simple pendulum has mass m and is oscillating with an amplitude a. If the length of the pendulum is L, then the maximum tension in the string is \left[\cos 0^{\circ}=1\right., g= acceleration due to gravity]

Choose the correct answer: