Question

A simple pendulum of length 'l' and a bob of mass 'm' is executing S.H.M. of small amplitude 'A'. The maximum tension in the string will be ($g =$ acceleration due to gravity)

• A. 2 mg
• B. $mg [1 + (\frac{A}{l})^2]$
• C. $mg [1 + (\frac{A}{l})]^2$
• D. $mg [1 + (\frac{A}{l})]$

Hint:

Physics Tip: Tension is always highest at the bottom of the swing because that is where the weight vector acts directly against the string and the speed (and thus centripetal force) is at its peak.

Answer 1

The Correct Option is B

Concept: Physics (Oscillations) - Tension in a Simple Pendulum.

Step 1: State the general formula for tension. The tension ($T'$) in the string at any angle $\theta$ is the sum of the radial component of weight and the required centripetal force: $$ T' = mg \cos \theta + \frac{mv^2}{l} $$

Step 2: Determine the conditions for maximum tension. Maximum tension occurs at the mean position where $\theta = 0$ (so $\cos \theta = 1$) and velocity is at its maximum ($v_{max}$). $$ T_{max}' = mg + \frac{mv_{max}^2}{l} $$

Step 3: Find maximum velocity in terms of amplitude. For S.H.M., $v_{max} = A\omega$. Since the angular frequency of a simple pendulum is $\omega = \sqrt{\frac{g}{l}}$: 1393] $$ v_{max}^2 = A^2 \left( \frac{g}{l} \right) $$

Step 4: Substitute and simplify. $$ T_{max}' = mg + \frac{m(A^2 g / l)}{l} = mg + \frac{mA^2 g}{l^2} $$ $$ T_{max}' = mg \left[ 1 + \frac{A^2}{l^2} \right] = mg \left[ 1 + \left( \frac{A}{l} \right)^2 \right] $$ $$ \therefore \text{The maximum tension is } mg \left[ 1 + \left( \frac{A}{l} \right)^2 \right]. $$