The 'period' of a pendulum . . . the time it takes to go back and forth once, and return to where it started . . . is
T = 2π √(length/gravity)
For this pendulum,
T = 2π √(0.24m / 9.8 m/s²)
T = 2π √0.1565 s²
T = 0.983 second
If you pull it to the side and let it go, it hits its highest speed at the BOTTOM of the swing, where all the potential energy you gave it has turned to kinetic energy. That's 1/4 of the way through a full back-and-forth cycle.
For this pendulum, that'll be (0.983s / 4) =
<em>(A). T = 0.246 second</em> <em><===</em>
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Notice that the formula T = 2π √(length/gravity) doesn't say anything about how far the pendulum is swinging. For small angles, it doesn't make any difference how far you pull it before you let it go . . . the period will be the same for tiny swings, little swings, and small swings. It doesn't change if you don't pull it away too far. So . . .
<em>(B).</em> The period is the same whether you pulled it 3.5 or 1.75 . <em>T = 0.246 s.</em>
Answer:
An nonlinear inverse relationship
Explanation:
C. located in front of the lens
