16.4 The simple pendulum  (Page 3/4)

A ball is attached to a string of length 4 m to make a pendulum. The pendulum is placed at a location that is away from the Earth’s surface by twice the radius of the Earth. What is the acceleration due to gravity at that height and what is the period of the oscillations?

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

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

Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.1 kg attached to it and has a length of 5 m. Pendulum 2 has a ball of mass 0.5 kg attached to a string of length 1 m. How does mass of the ball affect the frequency of the pendulum? Which pendulum will have a higher frequency and why?

Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.

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

Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?

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

How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

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

Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?

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

A pendulum with a period of 2.00000 s in one location $\left(g=9\text{.}\text{80}{\text{m/s}}^2\right)$ is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?

(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%?

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is $1\text{.}\text{63}{\text{m/s}}^2$ .

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

Suppose the length of a clock’s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.

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?