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Answer :
To find the length [tex]\( L \)[/tex] of the pendulum when the period [tex]\( T \)[/tex] is 1.57 seconds, we need to use the formula for the period of a pendulum:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{32}} \][/tex]
We are given:
- [tex]\( T = 1.57 \)[/tex] seconds
- [tex]\(\pi = 3.14\)[/tex]
Let's rearrange the formula to solve for [tex]\( L \)[/tex].
1. Start with the formula:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{32}} \][/tex]
2. To isolate the square root, divide both sides by [tex]\( 2 \pi \)[/tex]:
[tex]\[ \frac{T}{2 \pi} = \sqrt{\frac{L}{32}} \][/tex]
3. Square both sides to remove the square root:
[tex]\[ \left(\frac{T}{2 \pi}\right)^2 = \frac{L}{32} \][/tex]
4. Multiply both sides by 32 to solve for [tex]\( L \)[/tex]:
[tex]\[ L = 32 \left(\frac{T}{2 \pi}\right)^2 \][/tex]
Now, plug in the values:
- [tex]\( T = 1.57 \)[/tex]
- [tex]\(\pi = 3.14\)[/tex]
[tex]\[ L = 32 \left(\frac{1.57}{2 \times 3.14}\right)^2 \][/tex]
Calculating the above expression results in:
[tex]\[ L \approx 2.0 \text{ feet} \][/tex]
Therefore, the length [tex]\( L \)[/tex] of the pendulum that gives a period of 1.57 seconds is approximately 2 feet.
[tex]\[ T = 2 \pi \sqrt{\frac{L}{32}} \][/tex]
We are given:
- [tex]\( T = 1.57 \)[/tex] seconds
- [tex]\(\pi = 3.14\)[/tex]
Let's rearrange the formula to solve for [tex]\( L \)[/tex].
1. Start with the formula:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{32}} \][/tex]
2. To isolate the square root, divide both sides by [tex]\( 2 \pi \)[/tex]:
[tex]\[ \frac{T}{2 \pi} = \sqrt{\frac{L}{32}} \][/tex]
3. Square both sides to remove the square root:
[tex]\[ \left(\frac{T}{2 \pi}\right)^2 = \frac{L}{32} \][/tex]
4. Multiply both sides by 32 to solve for [tex]\( L \)[/tex]:
[tex]\[ L = 32 \left(\frac{T}{2 \pi}\right)^2 \][/tex]
Now, plug in the values:
- [tex]\( T = 1.57 \)[/tex]
- [tex]\(\pi = 3.14\)[/tex]
[tex]\[ L = 32 \left(\frac{1.57}{2 \times 3.14}\right)^2 \][/tex]
Calculating the above expression results in:
[tex]\[ L \approx 2.0 \text{ feet} \][/tex]
Therefore, the length [tex]\( L \)[/tex] of the pendulum that gives a period of 1.57 seconds is approximately 2 feet.
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