Question

A bob of a simple pendulum of mass $m$ is displaced through $90^\circ$ from its mean position and released. When the bob is at its lowest position, the tension in the string is

• A. $4\ mg$
• B. $2\ mg$
• C. $mg$
• D. $3\ mg$

Hint:

This is a standard scenario in circular motion and mechanics! For any object released from a horizontal position ($90^\circ$), the centripetal force at the bottom is always exactly $2mg$. Adding the rest weight of the object ($1mg$) to counter gravity gives a total vertical tension of exactly $3mg$. Memorizing this relationship can save valuable time during examinations.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A simple pendulum bob is pulled horizontally until the supporting string is perfectly level (displaced by $90^\circ$ from the vertical resting position) and let go from rest. We want to evaluate the maximum tension pulling along the string at the lowest vertical boundary point of its swing.

Step 2: Key Formula or Approach:
1. Use the Law of Conservation of Energy to find the velocity ($v$) of the bob at its lowest point. The loss in gravitational potential energy equals the gain in kinetic energy: $$mgh = \frac{1}{2}mv^2$$ 2. At the lowest position, the net vertical force acting on the bob provides the necessary centripetal acceleration ($a_c = \frac{v^2}{l}$) to sustain circular motion: $$T - mg = \frac{mv^2}{l}$$ Where $T$ is the string tension and $l$ is the length of the pendulum string.

Step 3: Detailed Explanation:
When the pendulum is released from an angle of $90^\circ$, it drops down by a vertical height equal to the full length of the string ($h = l$). Equating the potential energy loss to kinetic energy gain: $$mgl = \frac{1}{2}mv^2$$ Isolating the kinetic energy component $mv^2$: $$mv^2 = 2mgl \implies \frac{mv^2}{l} = 2mg$$ Now, write down the balance of forces at the bottom-most point. The tension $T$ pulls vertically upwards toward the center pivot, while gravity $mg$ pulls straight down: $$T - mg = \frac{mv^2}{l}$$ Substitute the centripetal force expression ($\frac{mv^2}{l} = 2mg$) into the equation: $$T - mg = 2mg$$ Isolating the tension variable $T$: $$T = 2mg + mg = 3mg$$

Step 4: Final Answer:
The tension in the string at the lowest position is $3mg$, which corresponds to option (D).