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The period T of a pendulum (the time it takes to make one complete oscillation) varies directly as the square root of its length L. A pendulum 7 ft long has a period of 2.9 seconds. find the period (in s) of a pendulum that is 10 ft long. round to the nearest tenth of a second. _____ swhat is the length (in ft) of a pendulum that beats seconds (that is, has a 2 second period)? round to the nearest tenth of a ft. ______ ft

Sagot :

The problem is about a directly proportional variation between T and the square root of L.

[tex]T=k\cdot\sqrt[]{L}[/tex]

Where k is the constant of proportionality. Let's use L = 7ft and T = 2.9 sec to find k.

[tex]2.9=k\cdot\sqrt[]{7}\to k=\frac{2.9}{\sqrt[]{7}}\approx1.10[/tex]

The constant of proportionality is 1.10.

Then, use the constant to find the period of a pendulum that is 10 feet long.

[tex]T=1.10\sqrt[]{10}\approx3.5\sec [/tex]

The period of a pendulum that is 10 ft long is 3.5 seconds.

Then, use T = 2 sec to find the length.

[tex]\begin{gathered} 2=1.10\sqrt[]{L} \\ L=(\frac{2}{1.10})^2 \\ L\approx3.3ft \end{gathered}[/tex]

The length of a pendulum that beats seconds is 3.3 feet.

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