Question 195: Oscillations

Question

A simple pendulum with length $$100 \mathrm{~cm}$$ and bob of mass $$250 \mathrm{~g}$$ is executing S.H.M. of amplitude $$10 \mathrm{~cm}$$. The maximum tension in the string is found to be $$\frac{x}{40} \mathrm{~N}$$. The value of $$x$$ is ___________.

Accepted Answer

Given the amplitude $$A$$ and the length $$l$$ of the pendulum, we can find the maximum angular displacement $$	heta_0$$:
$$
\sin 	heta_0 = \frac{A}{l} = \frac{10}{100} = \frac{1}{10}
$$
By conservation of energy, the following equation holds:
$$
\frac{1}{2} m v^2 = m g l(1 - \cos 	heta)
$$
The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:
$$
\begin{aligned}
& T - mg = \frac{m v^2}{l} \
& \Rightarrow T = mg + \frac{m v^2}{l}
\end{aligned}
$$
Substituting the conservation of energy equation, we get:
$$
\begin{aligned}
& T = mg + 2 m g(1 - \cos 	heta) \\
& = mg\left[1 + 2\left(1 - \sqrt{1 - \sin^2 	heta}ight)ight] \\
& = mg\left[3 - 2 \sqrt{1 - \frac{1}{100}}ight] \\
& = \frac{250}{1000} 	imes 10\left[3 - 2\left(1 - \frac{1}{200}ight)ight] = \frac{101}{40} \\
& 	herefore x = 101
\end{aligned}
$$
So, the value of $$x$$ is 101.

Oscillations

A simple pendulum with length $100 \mathrm{~cm} and bob of mass 250 \mathrm{~g} is executing S.H.M. of amplitude 10 \mathrm{~cm}. The maximum tension in the string is found to be \frac{x}{40} \mathrm{~N}. The value of x$ is ___________.

Choose the correct answer: