Question

If a simple pendulum of length 3.09 m and mass 8.9 kg is given an initial speed of 0.92 m/s at its equilibrium position, what is its period (assuming simple harmonic motion)?

a) 1.82 s
b) 2.15 s
c) 2.50 s
d) 3.06 s

Answer 1

To calculate the period of a simple pendulum, use the formula T = 2π√(L/g). For the given length of 3.09 m, the approximate period is 3.54 s, which is not one of the provided options, indicating a possible error in the options or the calculation.

Step-by-step explanation:

To determine the period of a pendulum in simple harmonic motion, we can use the formula for the period (T) of a simple pendulum, which is:

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, typically approximated as 9.81 m/s² on Earth's surface.

In the given question, the length (L) of the pendulum is 3.09 m and we assume g to be 9.81 m/s². Plugging in these values, we get:

T = 2π√(3.09/9.81)

Calculating this using a calculator:

T = 2π√(0.315) ≈ 2π√(0.561)≈ 2π(0.749) ≈ 4.712√(0.749) ≈ 4.712 × 0.865 ≈ 4.076

T ≈ 3.54 s (approx)

However, this value is not one of the options provided in the question, which suggests that there might be an error in the options or in the calculation. It is important to check that the length and the initial speed provided do not affect the period, as the period of a simple pendulum depends only on the length and local acceleration due to gravity.