Adjust pendulum
Problem
A time for one complete oscillation is a period $T$. The period for a simple pendulum is
given by $T = 2\pi\sqrt{\frac{l}{g}}$ where $l$ is the length of the pendulum and $g$ is the acceleration of gravity.
1) Calculate the period for a simple pendulum if $l = 30$ cm and $g = 9.81 m/s^2$.
This period is not equal to 1 second. There are few possibilities to modify the pendulum or change the conditions to get the pendulum with the period of 1 second.
2) First of all you can change the length of the pendulum. Find the required length.
3) It is possible to change the acceleration of gravity $g$ for example if you are in a lift which is accelerating. Find a new $g'$ if $l = 30$ cm as before. Is the lift moving up or down?
4) There is a bit more interesting way of modifying the pendulum.
Hammer a nail to the wall or make any other point obstacle for the thread and then swing the pendulum near the wall. Could you derive the period formula for this pendulum? Find $x$ in order to have the resultant period 1 second if $l = 30$cm and $g = 9.81 m/s^2$.
5) Could you think of any other way how to modify the pendulum?
Getting Started
Think about a car which starts to move really fast. What happens? Try to use analogy for the lift.
If you place an obstable for the thread you will have two pendulums of different length. Thus, the resultant period is a combination of periods for different lengths.
Student Solutions
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.