16.4 The Simple Pendulum

An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of $\text{10} \text{kg}$. Pendulum 2 has a bob with a mass of $\text{100 kg}$. Describe how the motion of the pendula will differ if the bobs are both displaced by $\text{12}º$.

Which of the following gives the correct relation between the acceleration due to gravity and period of a pendulum?

1. $g \textrm{ } = \textrm{ } \frac{2 \pi L}{T^{2}}$
2. $g \textrm{ } = \textrm{ } \frac{4 \pi^{2} L}{T^{2}}$
3. $g \textrm{ } = \textrm{ } \frac{2 \pi L}{T}$
4. $g \textrm{ } = \textrm{ } \frac{2 \pi^{2} L}{T}$

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

What is the length of a pendulum that has a period of 0.500 s?

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

What is the period of a 1.00-m-long pendulum?

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is $9 . \text{79} \text{m}/\text{s}^{2}$ is moved to a location where the acceleration due to gravity is $9 . \text{82} \text{m}/\text{s}^{2}$. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation…

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

(a) What is the effect on the period of a pendulum if you double its length?

(b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is $1 . \text{63} \text{m}/\text{s}^{2}$, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s minute hand, which on Earth makes one revolution per hour, to make one revolution on the Moon.

As usual, the acceleration due to gravity in these problems is taken to be $g = 9.80 \text{m} / \text{s}^{2}$, unless otherwise specified.

If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time?