# Find, to the nearest tenth of a second, the period of a pendulum of length 3.5 f

Find, to the nearest tenth of a second, the period of a pendulum of length 3.5 feet.

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Asma Vang
Given:
To calculated period of pendulums of the given lengths are 2 feet, 3 feet and 4,5 feet.
Coefficient of friction is 0.4
Formula: $$\displaystyle{T}={2}\pi\sqrt{{{\frac{{{L}}}{{{32}}}}}}$$
where L is the length of period of the pendulum.
Calculating the period of pendulum of length 2 feet :
$$\displaystyle{T}={2}{\left({3.14}\right)}\sqrt{{{\frac{{{2}}}{{{32}}}}}}$$
$$\displaystyle={2}{\left({3.14}\right)}\sqrt{{{0.0625}}}$$
Taking square root, $$\displaystyle{T}={2}{\left({3.14}\right)}{\left({0.25}\right)}$$
$$\displaystyle={1.57}$$
$$\displaystyle={1.6}$$ seconds
Calculating period of pendulum of length 3 feet.
$$\displaystyle{T}={2}{\left({3.14}\right)}\sqrt{{{\frac{{{3}}}{{{32}}}}}}$$
$$\displaystyle={2}{\left({3.14}\right)}{\left({0.31}\right)}$$
$$\displaystyle={1.9468}$$
$$\displaystyle={1.9}$$ seconds
Calculating period of pendulum of length 4.5 feet:
$$\displaystyle{t}={2}{\left({3.14}\right)}\sqrt{{{\frac{{{4.5}}}{{{3.2}}}}}}$$
$$\displaystyle={2}{\left({3.14}\right)}\sqrt{{{0.140625}}}$$
$$\displaystyle={2}{\left({3.14}\right)}{\left({0.375}\right)}$$
$$\displaystyle={2.355}$$
$$\displaystyle={2.4}$$ seconds
Therefore, the periods of pendulum of lengths 2 feet, 3 feet, 4.5 feet are 1.6 seconds, 1.9 seconds, and 2.4 seconds respectively.