Question

A simple pendulum of length I has a bob of mass m. It executes SHM of small amplitude A. The maximum tension in the string is (g = acceleration due to gravity)

Options

• mg

• ${\displaystyle \text{mg}(\frac{{\text{A}}^2}{{\text{l}}^2}+1)}$

• 2 mg

• ${\displaystyle \text{mg}(\frac{\text{A}}{\text{l}}+1)}$

Answer 1

${\displaystyle \text{mg}(\frac{{\text{A}}^2}{{\text{l}}^2}+1)}$

Explanation:

Consider the figure shown below

The string possess maximum tension when bob is at mean position of oscillation i.e., at position R.

From geometry, OP = ${\displaystyle \sqrt{{\text{l}}^2-{\text{A}}^2}}$

Also, RP = OR - OP = l - ${\displaystyle \sqrt{{\text{l}}^2-{\text{A}}^2}}$

The whole kinetic energy of bob at position R is converted into its potential energy at position B.

${\displaystyle \therefore \frac{1}{2}{\text{mv}}^2=\text{mg}(\text{l}-\sqrt{{\text{l}}^2-{\text{A}}^2})}$

${\displaystyle {\text{v}}^2=\text{2g}(\text{l}-\sqrt{{\text{l}}^2-{\text{A}}^2})}$

Balancing forces at R,

T - mg = ${\displaystyle \frac{{\text{mv}}^2}{\text{l}}=\frac{2\text{mg}(\text{l}-\sqrt{{\text{l}}^2-{\text{A}}^2})}{\text{l}}}$

∴ T = mg + 2mg ${\displaystyle (1-\sqrt{1-\frac{{\text{A}}^2}{{\text{l}}^2}})}$

Using approximation, ${\displaystyle \sqrt{1-x^2}=1-\frac{x^2}{2}}$ for x << 1, we get

T = mg = 2mg ${\displaystyle [1-(1-\frac{{\text{A}}^2}{2{\text{l}}^2})]}$

= mg + mg${\displaystyle {\left(\frac{\text{A}}{\text{l}}\right)}^2}$

= mg ${\displaystyle [1+{\left(\frac{\text{A}}{\text{l}}\right)}^2]}$