What change would double the period of a pendulum?

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Multiple Choice

What change would double the period of a pendulum?

Explanation:
The period of a simple pendulum, which is the time it takes to complete one full oscillation, is determined primarily by its length and the acceleration due to gravity. The formula for the period \( T \) of a pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. To double the period \( T \), we can derive the necessary change based on the formula. If we want \( T' = 2T \), we substitute into the equation to find out how \( L \) must change: \[ 2T = 2\pi \sqrt{\frac{L'}{g}} \] Squaring both sides gives: \[ (2T)^2 = 4T^2 = 4\pi^2 \frac{L'}{g} \] Since we know from the original formula that: \[ T^2 = \frac{4\pi^2 L}{g} \] we can set those equal: \[ 4T^2 = 4\pi^2 \frac{L

The period of a simple pendulum, which is the time it takes to complete one full oscillation, is determined primarily by its length and the acceleration due to gravity. The formula for the period ( T ) of a pendulum is given by:

[

T = 2\pi \sqrt{\frac{L}{g}}

]

where ( L ) is the length of the pendulum and ( g ) is the acceleration due to gravity.

To double the period ( T ), we can derive the necessary change based on the formula. If we want ( T' = 2T ), we substitute into the equation to find out how ( L ) must change:

[

2T = 2\pi \sqrt{\frac{L'}{g}}

]

Squaring both sides gives:

[

(2T)^2 = 4T^2 = 4\pi^2 \frac{L'}{g}

]

Since we know from the original formula that:

[

T^2 = \frac{4\pi^2 L}{g}

]

we can set those equal:

[

4T^2 = 4\pi^2 \frac{L

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