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Tarzan is swinging back and forth on a 6.25 m long vine. What is the period of his oscillation? (Unit = s)

Sagot :

To determine the period of oscillation of Tarzan swinging on a vine, we can treat this scenario as a simple pendulum. The period of a simple pendulum is given by the formula:

[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]

where:
- [tex]\( T \)[/tex] is the period of oscillation,
- [tex]\( L \)[/tex] is the length of the vine (which in this case acts as the length of the pendulum),
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex].

Given data:
- The length of the vine [tex]\( L = 6.25 \, \text{m} \)[/tex],
- The acceleration due to gravity [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex].

Let's break down the steps to find the period [tex]\( T \)[/tex]:

1. Divide the length of the vine by the acceleration due to gravity:
[tex]\[ \frac{L}{g} = \frac{6.25}{9.81} \][/tex]

2. Take the square root of the result from step 1:
[tex]\[ \sqrt{\frac{L}{g}} = \sqrt{\frac{6.25}{9.81}} \][/tex]

3. Multiply the result by [tex]\( 2 \pi \)[/tex] to find the period [tex]\( T \)[/tex]:
[tex]\[ T = 2 \pi \sqrt{\frac{6.25}{9.81}} \][/tex]

Upon evaluating this expression, the final period of oscillation [tex]\( T \)[/tex] is:

[tex]\[ T \approx 5.015 \, \text{s} \][/tex]

Thus, the period of Tarzan's oscillation on the 6.25 m long vine is approximately [tex]\( 5.015 \)[/tex] seconds.