Question

The angular amplitude of a simple pendulum is $\theta_{0}$. The maximum tension in its string will be

• A. $m g\left(1-\theta_{0}\right)$
• B. $m g\left(1+\theta_{0}\right)$
• C. $m g\left(1-\theta_{0}^{2}\right)$
• D. $m g\left(1+\theta_{0}^{2}\right)$

Answer 1

The Correct Option is D

The simple pendulum at angular amplitude $\theta_{0}$ is shown in the figure. Maximum tension in the string is
$T_{\max }=m g+\frac{m v^{2}}{l}\,\,\,...(i)$
When bob of the pendulum comes from $A$ to $B$, it covers a vertidal distance $h$

$\therefore \cos \theta_{0}=\frac{l-h}{l}$
$\Rightarrow h=l\left(1-\cos \theta_{0}\right)\,\,\,...(ii)$
Also during $A$ to $B$, potential energy of bob converts into kinetic energy ie, $m g h=\frac{1}{2} m v^{2}$
$\therefore v=\sqrt{2 g h}\,\,\,...(iii)$
Thus, using Eqs. (i), (ii) and (iii), we obtain
$T_{\max } =m g+\frac{2 m g}{l} l\left(1-\cos \theta_{0}\right) $
$=m g+2 m g\left[1-1+\frac{\theta_{0}^{2}}{2}\right] $
$=m g\left(1+\theta_{0}^{2}\right)$

Answer 2

Let's derive the expression for the maximum tension in the string of a simple pendulum with an angular amplitude \(\theta_0\).
Step 1: Small Angle Approximation
For small angular displacements, we use the small angle approximation:
\[ \cos\theta_0 \approx 1 - \frac{\theta_0^2}{2} \]
Step 2: Potential Energy at Maximum Displacement
The potential energy at maximum displacement is given by:
\[ PE = mgL(1 - \cos\theta_0) \]
Using the small angle approximation:
\[ PE = mgL \left(1 - \left(1 - \frac{\theta_0^2}{2}\right)\right) \]
\[ PE = mgL \cdot \frac{\theta_0^2}{2} \]
Step 3: Kinetic Energy at the Lowest Point
This potential energy converts to kinetic energy at the lowest point:
\[ KE = \frac{1}{2}mv^2 \]
Equating the potential and kinetic energy:
\[ mgL \cdot \frac{\theta_0^2}{2} = \frac{1}{2}mv^2 \]
Solving for \( v^2 \):
\[ v^2 = gL\theta_0^2 \]
Step 4: Centripetal Force at the Lowest Point
At the lowest point, the centripetal force required is:
\[ F_{\text{centripetal}} = \frac{mv^2}{L} \]
Substituting \( v^2 \):
\[ F_{\text{centripetal}} = \frac{m \cdot gL\theta_0^2}{L} \]
\[ F_{\text{centripetal}} = mg\theta_0^2 \]
Step 5: Total Tension in the String
The total tension \(T\) in the string at the lowest point is the sum of the gravitational force and the centripetal force:
\[ T = mg + mg\theta_0^2 \]
\[ T = mg(1 + \theta_0^2) \]
Conclusion
The maximum tension in the string of a simple pendulum with an angular amplitude \(\theta_0\) is:
\[ T = mg(1 + \theta_0^2) \]