A simple pendulum has mass 1.14 kg and length 0.760 m.

(a) What is the period of the pendulum near the surface of Earth?
(b) If the same mass is attached to a spring, what spring constant would result in the period of motion found in part (a)?

a. T^2 = 4pi^2(L/g)=4*3.14^2(0.76/9.8) =

3.06
T = 1.75 s.

To determine the period of a simple pendulum near the surface of the Earth, you 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.

(a) To find the period of the pendulum, we can substitute the values into the formula:

T = 2π * √(0.760 m / 9.8 m/s^2)

Calculating this, we have:

T ≈ 2π * √0.0776 ≈ 2π * 0.2787 ≈ 1.752 seconds

So, the period of the pendulum near the surface of the Earth is approximately 1.752 seconds.

(b) Now, let's consider the spring-mass system. We want to find the spring constant, which relates to the period of motion found in part (a).

The formula for the period of motion of a mass-spring system is:

T = 2π * √(m / k)

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

Since we want the same period of motion as in part (a), we can equate the two formulas:

2π * √(0.760 m / 9.8 m/s^2) = 2π * √(1.14 kg / k)

Simplifying this equation, we find:

√(0.760 / 9.8) = √(1.14 / k)

Squaring both sides, we have:

0.760 / 9.8 = 1.14 / k

Cross-multiplying, we get:

0.760k = 1.14 * 9.8

Dividing both sides by 0.760, we find:

k ≈ 15.0 N/m

Therefore, a spring constant of approximately 15.0 N/m would result in the same period of motion as calculated in part (a).