Question

The length of a simple pendulum is 0.65 mm , the pendulum bob has a mass of 295 gg , and it is released at an angle of 12 ∘∘ to the vertical. Assume SHM.With what frequency does it oscillate?

Express your answer to two significant figures and include the appropriate units.

Answer 1

The frequency of the pendulum can be calculated using the formula f = 1 / T, where T is the period of the pendulum. The period of a simple pendulum is given by T = 2π√(L/g). By substituting the given values, the period is calculated to be approximately 0.407 seconds. Finally, the frequency is calculated to be 2.45 Hz.

Step-by-step explanation:

The frequency of a simple pendulum can be calculated using the formula:

f = 1 / T

where T is the period of the pendulum.

The period of a simple pendulum is given by:

T = 2π√(L/g)

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

Using the given values:

L = 0.65 mm = 0.065 m

m = 295 gg = 0.295 kg

θ = 12°

Converting degrees to radians:

θ = 12° = 12 * (π/180) rad

Substituting the values into the formula:

T = 2π√(L/g) = 2π√(0.065/9.8) ≈ 0.407 s

Finally, calculating the frequency:

f = 1 / T = 1 / 0.407 = 2.45 Hz