Question

The bob of a simple pendulum weighs 25 g. The tension in the cord is 20 g-wt at a certain
(15 g-wt)
position when the pendulum is vibrating. Find the restoring force.

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Answer 1

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Answer 2

Concept:

A force that acts to return a body to its equilibrium position is known as the restorative force. The restoring force is always directed back toward the equilibrium position of the system and depends only on the position of the mass or particle.

An action-reaction pair of forces acting at each end of the string is tension.

Given:

The weight of the bob of a simple pendulum is $25g$.

The tension in the cord is $20\text{g-wt}$.

Find:

The restoring force.

Solution:

In a simple pendulum, the different forces acting on the bob and the cord are tension and restoring forces.

Now,

$T=mgcos(\theta)\\20\times 10=25\times 10(cos\theta)\\cos(\theta)=\frac{4}{5}$

Now since $sin^2(\theta)+cos^2(\theta)=1$

$sin(\theta)=\frac{3}{5}$

Therefore, the restoring force:

$F=-mgsin(\theta)\\F=25\times \frac{3}{5}$

$F=15\text{g-wt}$

The restoring force is $15\text{g-wt}$.

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