Final answer:
The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.
Step-by-step explanation:
(a) What length of pendulum has a period of 1.2 s on Earth? cm
Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.
Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:
L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm
Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:
L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.
Given that the spring constant is 10 N/m, substituting the values into the equation, we have:
m = (1.2/2π)² * 10 = 0.722 kg.
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg
Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:
m = (1.2/2π)² * 10 = 0.329 kg.