1. Misy pendule tsotra roa eo amin'ny toerana roa samy hafa. Ny halavan'ny pendule faharoa dia in-0.4 heny noho ny halavan'ny pendule voalohany, ary ny hainganan ny hery misintona niainan'ny Ny pendulum faharoa dia in-0.9 ny hafainganan'ny hery misintona niainan'ny ny pendulum voalohany. Fantaro ny comparison of ny matetika ny ny voalohany pendulum ny segondrad pendulum.
A. 2/3
B. 3/2
T. 4/9
D. 9/4
Fantatra:
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
Mitady: The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2)
vahaolana:
The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2):
Ny valiny marina dia A.
2. An object is suspended from iray end of a cord ary avy eo manao a fihetsiketsehana mirindra tsotra 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
Fantatra:
Frequency of pendulum (f) = 0.5 Hz
Mitady: Determine the period (T) of the pendulum if the length of cord (L) is four times the initial length
vahaolana:
fe-potoana 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 segondra.
Ny valiny marina dia D.
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 sy f2.
A. f1 = f2
B. f1 = 2 f2
C. f2 = 2 f1
D. f1 = 4 f2
vahaolana:
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 faobe does not affect the frequency of the simple pendulum.
Ny valiny marina dia A.
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
vahaolana:
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 tehina (l) and the acceleration of gravity (g) impact the period of the simple pendulum. Otherwise, the mass of ny zavatra ary ny voalohany zoro does not impact the fe-potoana simple pendulum.
Ny valiny marina dia 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 metatra
B. 0.50 metatra
C. 2.0 metatra
D. 4.0 metatra
Fantatra:
The mass does not impact the frequency of the simple pendulum.
The length of the cord of the simple pendulum (l) = 1 meter
Mitady: determine the length of rope if the frequency is twice the initial frequency
vahaolana:
The initial frequency of the simple pendulum :
The frequency of the simple pendulum is twice the initial frequency :
Fa ny farany frequency to be doubled, the length of the pendulum should be changed to 0.25 meters.
Ny valiny marina dia A.
