# Simple pendulum – problems and solutions

1. Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulum, and the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum. Determine the comparison of the frequency of the first pendulum to the second pendulum.

A. 2/3

B. 3/2

C. 4/9

D. 9/4

Known :

The length of the cord of the first pendulum (l1) = 1

The length of cord of the second pendulum (l2) = 0.4 (l1) = 0.4 (1) = 0.4

Acceleration due to the gravity of the first pendulum (g1) = 1

Acceleration due to gravity of the second pendulum (g2) = 0.9 (1) = 0.9

Wanted: The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2)

Solution :

The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2) :

2. An object is suspended from one end of a cord and then perform a simple harmonic motion with a frequency of 0.5 Hertz. If the length of the cord is increased by four times the initial length, then determine the period of the harmonic motion.

A. ¼ seconds

B. ½ seconds

C. 2 seconds

D. 4 seconds

Known :

Frequency of pendulum (f) = 0.5 Hz

Wanted: Determine the period (T) of the pendulum if the length of cord (l) is four times the initial length

Solution :

Period of the first pendulum :

The initial length of cord :

If the length of the cord is increased by four times the initial length :

Then the period of a pendulum is :

The period of motion is 4 seconds.

3. Two pendulums with the same length of its cord, but the mass of the second pendulum is four times the mass of the first pendulum. If f1 is the frequency of the first pendulum and f2 is the frequency of the second pendulum, then determine the relationship between f1 and f2.

A. f1 = f2

B. f1 = 2 f2

C. f2 = 2 f1

D. f1 = 4 f2

Solution :

The equation of frequency of the simple pendulum :

f = frequency, g = acceleration due to gravity, l = the length of cord

Based on the equation above, can conclude that mass does not affect the frequency of the simple pendulum.

4. The quantities below that do not impact the period of the simple pendulum are…..

A. length of cord and mass of the object

B. length of cord and acceleration due to gravity

C. mass of the object and initial angle

D. length of cord and initial angle

Solution :

The equation of period of the simple pendulum :

T = period, g = acceleration due to gravity, l = length of cord

Based on the above formula, can conclude the length of the rod (l) and the acceleration of gravity (g) impact the period of the simple pendulum. Otherwise, the mass of the object and the initial angle does not impact the period of the simple pendulum.

5. The rope of the simple pendulum made from nylon. At one end of the rope suspended a mass of 10 gram and length of rope is 1 meter. If the frequency produced twice the initial frequency, then the length of the rope must be changed to…

A. 0.25 meters

B. 0.50 meters

C. 2.0 meters

D. 4.0 meters

Known :

The mass does not impact the frequency of the simple pendulum.

The length of the cord of the simple pendulum (l) = 1 meter

Wanted: determine the length of rope if the frequency is twice the initial frequency

Solution :

The initial frequency of the simple pendulum :

The frequency of the simple pendulum is twice the initial frequency :

For the final frequency to be doubled, the length of the pendulum should be changed to 0.25 meters.