When a simple pendulum swings, the tension in the string is not just balancing the component of the gravitational force but also provides the necessary centripetal force for the circular motion of the bob.

Forces Acting on the Pendulum Bob:

At any angular displacement $ \theta $:

Gravitational Force ($ mg $): Acts downward.

Tension in the String ($ T $): Acts along the string towards the pivot.

Radial Component of Gravitational Force: $ mg \cos \theta $

Tangential Component of Gravitational Force: $ mg \sin \theta $

Net Radial Force:

The net radial force provides the centripetal acceleration required for circular motion:

$ T - mg \cos \theta = m \frac{v^2}{L} $

Where:

$ T $ is the tension in the string,

$ v $ is the speed of the bob at angle $ \theta $,

$ L $ is the length of the pendulum.

Calculating the Speed $ v $ at $ \theta = 20^\circ $:

Using conservation of mechanical energy between the highest point ($ \theta_{\text{max}} = 40^\circ $) and the point at $ \theta = 20^\circ $:

Potential Energy at $ \theta_{\text{max}} $:

$ U_{\text{max}} = mgL (1 - \cos 40^\circ) $

Potential Energy at $ \theta = 20^\circ $:

$ U = mgL (1 - \cos 20^\circ) $

Kinetic Energy at $ \theta = 20^\circ $:

$ K = \frac{1}{2} m v^2 = U_{\text{max}} - U $

Compute the difference in potential energy:

$ \Delta U = mgL (\cos 20^\circ - \cos 40^\circ) $

Using known values:

$ \cos 20^\circ \approx 0.9397 $

$ \cos 40^\circ \approx 0.7660 $

Calculate $ \Delta U $:

$ \Delta U = mgL (0.9397 - 0.7660) = mgL (0.1737) $

Set $ \Delta U = \frac{1}{2} m v^2 $ and solve for $ v^2 $:

$ \frac{1}{2} m v^2 = mgL (0.1737) \implies v^2 = 2gL (0.1737) $

Calculating the Centripetal Acceleration:

$ a_{\text{centripetal}} = \frac{v^2}{L} = \frac{2gL (0.1737)}{L} = 2g (0.1737) = 0.3474g $

Calculating the Tension $ T $:

$ T = mg \cos \theta + m a_{\text{centripetal}} = mg \cos 20^\circ + m (0.3474g) $

Plugging in $ \cos 20^\circ $:

$ T = mg (0.9397) + m (0.3474g) = mg (0.9397 + 0.3474) = mg (1.2871) $

Conclusion:

The tension $ T $ is greater than $ mg \cos 20^\circ $ because it includes the additional centripetal force component.

Therefore, at an angular displacement of $ 20^\circ $, the tension in the string is indeed greater than $ mg \cos 20^\circ $.

Therefore, the statement is TRUE.