The angular amplitude of a simple pendulum is θ₀. The maximum tension in its string will be

mg (1 - θ₀)

mg (1 + θ₀ )

$mg (1 - θ_{0}^{2})$

$mg (1 + θ_{0}^{2})$

Solution

$mg (1 + θ_{0}^{2})$

The simple pendulum at angular amplitude θ_o is shown in the figure

T_max = $mg + \frac{{mv}^{2}}{l}$ ....(i)

When bob of pendulum comes from A to B, it covers a vertical distance h.

∴ ${cosθ}_{0} = \frac{l - h}{l}$

⇒ h = l ( l - cos θ ) .....(ii)

Also during A to B, potential energy of bob converts into kinetic energy

∴ mgh = $\frac{1}{2} {mv}^{2}$

∴ v = $\sqrt{2 g h}$ ....... (iii)

Thus, using Eqs. (i), (ii) and (iii) we obtain

T_max = mg + $\frac{2 mg}{l} l (1 - {cosθ}_{o})$

= mg + 2mg $[1 - 1 + \frac{θ_{0}^{2}}{2}]$

(By using binomial theorem)

= mg $(1 + θ_{0}^{2})$

Some More Questions From Oscillations Chapter

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