Question

A certain simple pendulum has a period on earth of 1.60 s. what is its period on the surface of Mars , where the acceleration due to gravity is 3.71 m/s2?

Answer 1

The period of a simple pendulum on the surface of Mars can be calculated using the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Step-by-step explanation:

To find the period of a simple pendulum on the surface of Mars, we can use the formula:

T = 2π√(L/g)

Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. Since we know the period on Earth (1.60 s) and the acceleration due to gravity on Mars (3.71 m/s^2), we can set up the equation:

1.60 s = 2π√(L/3.71 m/s²)

Solving for L gives:

L = 5.77 m

Therefore, the period of the pendulum on the surface of Mars is 1.60 seconds.

Answer 2

The period of a pendulum on Mars, where gravity is weaker than on Earth, will be longer. Using the formula T2 = T1 * √(g1/g2), the period on Mars is calculated to be approximately 2.60 seconds for a pendulum that has a period of 1.60 seconds on Earth.

Step-by-step explanation:

To find the new period, first, you can express the ratio of the periods at two different gravitational accelerations (T2/T1 = √(g1/g2)), and then solve for the new period T2 using the known period on Earth (T1).

If we let T1 = 1.60 s be the period on Earth and g1 = 9.81 m/s² the acceleration due to gravity on Earth, and g2 = 3.71 m/s² the acceleration on Mars, then:

T2 = T1 * √(g1/g2)
T2 = 1.60 s * √(9.81/3.71)
T2 = 1.60 s * √(2.64)
T2 = 1.60 s * 1.625
T2 ≈ 2.60 s

Therefore, the period of the certain pendulum on the surface of Mars is approximately 2.60 seconds.