The free-fall acceleration on Mars is 3.7 m/s2.

(a) What length pendulum has a period of 4 s on the earth?
cm
What length pendulum would have a 4 s period on Mars?
cm
(b) A mass is suspended from a spring with force constant 10 N/m. Find the mass suspended from this spring that would result in a 4 s period on Earth and on Mars.
Earth kg
Mars kg

The period of a pendulum is

2*pi*sqrt(L/g)

The period of a spring/mass system is
P = 2*pi*sqrt(m/k)

(k is the spring constant and you can guess what m is)

Use these formulas to solve your problems.

To find the length of a pendulum, you can use the formula:

T = 2π√(L/g)

where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.

(a) Length of pendulum with a 4 s period on Earth:
On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. So, to find the length of a pendulum with a 4 s period on Earth, we can rearrange the formula:

T = 2π√(L/g)

4 = 2π√(L/9.8)

Simplifying the equation, we get:

2 = π√(L/9.8)

2/π = √(L/9.8)

Squaring both sides, we have:

(2/π)^2 = L/9.8

L = (2/π)^2 * 9.8

Calculating the length, the pendulum on Earth would be approximately 1.0 meters.

(b) Length of pendulum with a 4 s period on Mars:
On Mars, the acceleration due to gravity is given as 3.7 m/s^2. Using the same formula as before:

T = 2π√(L/g)

4 = 2π√(L/3.7)

Simplifying the equation:

2 = π√(L/3.7)

2/π = √(L/3.7)

Squaring both sides:

(2/π)^2 = L/3.7

L = (2/π)^2 * 3.7

Calculating the length, the pendulum on Mars would be approximately 0.3741 meters or 37.41 centimeters.

For part (b), we need to use Hooke's Law for a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the force constant of the spring.

For Earth:
We want to find the mass that results in a 4 s period on Earth. Using the formula:

T = 2π√(m/k)

4 = 2π√(m/10)

Simplifying:

2 = π√(m/10)

2/π = √(m/10)

Squaring both sides:

(2/π)^2 = m/10

m = 10 * (2/π)^2

Calculating the mass, it would be approximately 1.28 kilograms.

For Mars:
Using the same formula as before:

T = 2π√(m/k)

4 = 2π√(m/10)

Simplifying:

2 = π√(m/10)

2/π = √(m/10)

Squaring both sides:

(2/π)^2 = m/10

m = 10 * (2/π)^2

Calculating the mass, it would still be approximately 1.28 kilograms. The mass suspended from this spring that would result in a 4 s period is the same for both Earth and Mars.