1. Две едноставни нишала се наоѓаат на две различни места. Должината на второто нишало е 0.4 пати поголема од должината на првото нишало, а забрзувањеn од гравитацијата искусни од страна на Второто нишало е 0.9 пати поголемо од забрзувањето на гравитацијата искусни од првото нишало. Определи го cспоредба на на фреквенцијата на на првата нишало до секундаd нишало.
А. 2/3
Б. 3/2
В. 4/9
Д. 9/4
Познато:
The length of the cord of the first pendulum (l1) = 1
The length of cord of the second pendulum (l2) = 0.4 (л1) = 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
Сакаше: The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2)
Решение:
The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2):
Точен одговор е А.
2. An object is suspended from една end of a cord и потоа изврши а едноставно хармонично движење 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
Околу 2 секунди
Д. 4 секунди
Познато:
Frequency of pendulum (f) = 0.5 Hz
Сакаше: Determine the period (Т) of the pendulum if the length of cord (l) is four times the initial length
Решение:
период 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 секунди.
Точниот одговор е Д.
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 f2.
A. f1 = ѓ2
B. f1 = 2 f2
C. f2 = 2 f1
D. f1 = 4 f2
Решение:
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 маса does not affect the frequency of the simple pendulum.
Точен одговор е А.
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
Решение:
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 прачка (l) and the acceleration of gravity (g) impact the period of the simple pendulum. Otherwise, the mass of објектот и почетната агол does not impact the период од simple pendulum.
Точниот одговор е C.
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 метри
B. 0.50 метри
C. 2.0 метри
D. 4.0 метри
Познато:
The mass does not impact the frequency of the simple pendulum.
The length of the cord of the simple pendulum (l) = 1 meter
Сакаше: determine the length of rope if the frequency is twice the initial frequency
Решение:
The initial frequency of the simple pendulum :
The frequency of the simple pendulum is twice the initial frequency :
на на финалето frequency to be doubled, the length of the pendulum should be changed to 0.25 meters.
Точен одговор е А.
