A simple pendulum with a bob of mass $m$ swings with an angular amplitude of $40^\circ$. When its angular displacement is $20^\circ$, the tension in the string is greater than $mg\cos 20^\circ$.

Step 1: Identify the parameters of the pendulum.

Let the simple pendulum have a bob of mass $m$ and a string of length $L$.

The angular amplitude is:

$$\theta_0 = 40^\circ$$

The instant of interest is at the angular displacement:

$$\theta = 20^\circ$$

Step 2: Analyze the forces acting on the bob.

At any angular displacement $\theta$, the forces acting on the bob are:

1. The tension $T$ acting along the string towards the pivot.
2. The gravitational force $mg$ acting vertically downwards.

Resolving the gravitational force $mg$ into components along and perpendicular to the string gives:

• Radial component: $mg\cos\theta$ (acting radially outwards)
• Tangential component: $mg\sin\theta$ (acting tangentially)

Step 3: Apply Newton's Second Law in the radial direction.

The net force towards the center of the circular path provides the centripetal acceleration.

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

Solve for the tension $T$.

$$T = mg\cos\theta + \frac{mv^2}{L}$$

Step 4: Determine the speed of the bob at $\theta = 20^\circ$.

Since the pendulum swings between $-40^\circ$ and $+40^\circ$, the bob only comes to rest at the extreme positions. At $\theta = 20^\circ$, the bob is in motion.

$$v > 0$$

This implies that the centripetal term is strictly positive.

$$\frac{mv^2}{L} > 0$$

Therefore, the tension must be greater than the radial component of gravity.

$$T > mg\cos 20^\circ$$

Step 5: Calculate the exact values using conservation of energy.

Using conservation of mechanical energy from the release point $\theta_0 = 40^\circ$ to the displacement $\theta$:

$$v^2 = 2gL(\cos\theta - \cos 40^\circ)$$

Substitute $v^2$ back into the expression for tension $T$.

$$T = mg\cos\theta + 2mg(\cos\theta - \cos 40^\circ)$$ $$T = mg(3\cos\theta - 2\cos 40^\circ)$$

Substitute $\theta = 20^\circ$ to find the numerical value of the tension.

$$T = mg(3\cos 20^\circ - 2\cos 40^\circ)$$ $$T \approx 1.29\,mg$$

Compare this with the value of $mg\cos 20^\circ$.

$$mg\cos 20^\circ \approx 0.94\,mg$$

Since $1.29\,mg > 0.94\,mg$, the tension is indeed greater than $mg\cos 20^\circ$.

Answer: The statement is True. At an angular displacement of $20^\circ$, the tension in the string is $T \approx 1.29\,mg$, which is greater than $mg\cos 20^\circ \approx 0.94\,mg$.