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Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

1) The period for the simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$. If $l = 30  cm = 0.3  m$ and $g = 9.81  m/s^2$ then $T = 2\pi\sqrt{\frac{0.3}{9.81}} = 1.1  s$.

2) Suppose a new length of the pendulum is $l'$. We have that $T = 1 s$ and $g = 9.81 m/s^2$.

Thus, $$l' = \frac{gT^2}{4\pi^2} = \frac{9.81}{4\pi^2} = 24.8  cm$$

3) Let $g'$ be a new acceleration of gravity then

$$g' = \frac{4\pi^2l}{T^2} = \frac{0.3\times 4\pi^2}{1^2} =11.8 m/s^2$$ $a = g' - g = 2.03 m/s^2$. So, it is a lift moving up with the acceleration of $2.03 m/s^2$. We know that it is moving up because $g' > g$. We are feeling heavier.

4) This time half of the period pendulum has the length $l$ and another half it has the $l - x$.

So, the resultant period $$T = \frac{T_1 + T_2}{2} = \frac{2\pi\sqrt{\frac{l}{g}} + 2\pi\sqrt{\frac{l - x}{g}}}{2} = \frac{\pi}{\sqrt{g}}(\sqrt{l} + \sqrt{l-x}) $$ Hence, we can calculate $x$. $$x = \frac{2T\sqrt{lg}}{\pi} - \frac{gT^2}{\pi^2} = \frac{2\sqrt{0.3\times 9.81}}{\pi} - \frac{9.81}{\pi^2} = 9.8  cm$$

It is possible to change the period changing the acceleration of gravity for example when you are accelerating in the car.