Question

Период колебаний математического маятника вычисляется по формуле T= , где T измеряется в секундах, - длина нити в метрах, - ускорение свободного падения в м/сек^2 . На Земле прием = 10 в м/сек^2 , на луне = 1.6 в м/сек^2. Маятник на Земле имеет длину 1 метр. Какова должна быть длина маятника на луне, чтобы их периоды совпадали? Ответ дайте в метрах.

Answer 1

T^2=4П^2l/g
4П^2l1/g1=4П^2l2/g2
l1/g1=l2/g2
l1g2=l2g1
l2=l1g2/g1=1*1.6/10=0.16
Ответ:0.16

Answer 2

Calculation of the Period of a Pendulum

The period of a mathematical pendulum can be calculated using the formula T = 2π√(L/g), where T is the period in seconds, L is the length of the pendulum in meters, and g is the acceleration due to gravity in m/s^2.

Length of the Pendulum on Earth

On Earth, the acceleration due to gravity is approximately 10 m/s^2 Let's assume the length of the pendulum on Earth is 1 meter.

Using the formula T = 2π√(L/g), we can calculate the period of the pendulum on Earth:

T = 2π√(1/10) T ≈ 2π√(0.1) T ≈ 2π * 0.316227766 T ≈ 1.989 seconds

Length of the Pendulum on the Moon

On the Moon, the acceleration due to gravity is approximately 1.6 m/s^2 We need to find the length of the pendulum on the Moon that will result in the same period as on Earth.

Let's assume the length of the pendulum on the Moon is L_m.

Using the formula T = 2π√(L/g), we can equate the periods on Earth and the Moon:

2π√(1/10) = 2π√(L_m/1.6)

Simplifying the equation:

√(1/10) = √(L_m/1.6)

1/√10 = √(L_m/1.6)

Squaring both sides of the equation:

1/10 = L_m/1.6

Cross-multiplying:

L_m = (1/10) * 1.6 L_m = 0.16 meters

Therefore, the length of the pendulum on the Moon should be approximately 0.16 meters in order for the periods to be the same.

Note: The above calculations assume ideal conditions and do not take into account factors such as air resistance or other environmental factors that may affect the motion of the pendulum.