Paprasta švytuoklė – problemos ir sprendimai

1. Dvi paprastos švytuoklės yra dviejose skirtingose ​​vietose. Antrosios švytuoklės ilgis yra 0.4 karto didesnis už pirmosios švytuoklės ilgį, o pagreitisn iš sunkis patyrė antrosios švytuoklės pagreitis yra 0.9 karto didesnis už gravitacijos pagreitį patyrė pirmoji švytuoklė. Nustatykite cpalyginimas As dažnis pirmas švytuoklė į sekundėd švytuoklė.

A. 2/3

B. 3/2

C. 4/9

D. 9/4

Žinomas:

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

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

sprendimas:

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

Teisingas atsakymas yra A.

2. An object is suspended from vienas end of a cord ir tada atlikti a paprastas harmoninis judesys 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

Apie 2 sekundžių

D. 4 sekundžių

Žinomas:

Frequency of pendulum (f) = 0.5 Hz

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

sprendimas:

laikotarpis 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 sekundžių.

Teisingas atsakymas yra 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 bei f₂.

A. f₁ = f₂

B. f₁ = 2 pėdų₂

C. f₂ = 2 pėdų₁

D. f₁ = 4 pėdų₂

sprendimas:

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 masė does not affect the frequency of the simple pendulum.

Teisingas atsakymas yra 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

sprendimas:

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 strypas (l) and the acceleration of gravity (g) impact the period of the simple pendulum. Otherwise, the mass of objektas ir pradinis kampas does not impact the laikotarpis simple pendulum.

Teisingas atsakymas yra 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 metro

B. 0.50 metro

C. 2.0 metro

D. 4.0 metro

Žinomas:

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

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

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

sprendimas:

The initial frequency of the simple pendulum :

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

Dėl As galutinis frequency to be doubled, the length of the pendulum should be changed to 0.25 meters.

Teisingas atsakymas yra A.