Question

a simple pendulum having a length of 1.65 m and a mass of 5.69 kg undergoes simple harmonic motion when given an initial speed of 1.69 m/s at its equilibrium position. determine its period. the acceleration due to gravity is 9.8 m/s 2 . answer in units of s.

Answer 1

The period of a simple pendulum with a length of 1.65 m and undergoing simple harmonic motion in a gravity field of 9.8 m/s² is approximately 2.5768 seconds.

Step-by-step explanation:

To determine the period of a simple pendulum undergoing simple harmonic motion, we use the formula for the period (T) of a pendulum:

T = 2π√(L/g)

Where:

• L is the length of the pendulum (in meters)
• g is the acceleration due to gravity (in meters per second squared)

In this case, L is 1.65 m and g is 9.8 m/s2. Inserting these values into the equation provides us with the period:

T = 2π√(1.65/9.8)

Calculating the square root and the multiplication by 2π gives us the period of the pendulum in seconds:

T = 2π√(0.16836735...) ≈ 2π(0.410325...) ≈ 2.5768 s

Therefore, the period of the pendulum is approximately 2.5768 seconds.