Question:

Consider a simple pendulum consisting of a mass of 100 g attached to a string of length 1 m. The pendulum is set into motion with an initial displacement of 10 degrees from the vertical and released from rest.

a) Calculate the angular frequency (ω) and period (T) of the pendulum.

b) Determine the maximum angular speed (ω_max) and maximum linear speed (v_max) of the pendulum during its motion.

c) Describe how the period and maximum angular amplitude of the pendulum would change if the length of the string is doubled.

Answer:

a) The angular frequency (ω) and period (T) of a simple pendulum can be calculated using the equation:

ω = √(g / L) and T = 2π / ω

Where:

• g is the acceleration due to gravity (9.8 m/s²)
• L is the length of the pendulum string

Given that the length of the pendulum is 1 m, we can calculate the angular frequency:

ω = √(9.8 m/s² / 1 m) ω = √9.8 rad/s ω ≈ 3.13 rad/s

Next, we can calculate the period of the pendulum:

T = 2π / ω T = 2π / 3.13 rad/s T ≈ 2.01 s

Therefore, the angular frequency (ω) of the pendulum is approximately 3.13 rad/s and the period (T) is approximately 2.01 seconds.

b) The maximum angular speed (ω_max) of the pendulum occurs at the equilibrium position (vertical). At this point, the pendulum is moving with its maximum speed. This angular speed can be calculated using the equation:

ω_max = √(2gh_max / L)

Where:

• h_max is the maximum height reached by the pendulum bob

Since the pendulum is released from rest, the maximum height reached is equal to the initial displacement.

h_max = L * cosθ h_max = 1 m * cos(10°) h_max ≈ 0.984 m

Now we can calculate the maximum angular speed:

ω_max = √(2 * 9.8 m/s² * 0.984 m / 1 m) ω_max = √(19.0968 rad²/s²) ω_max ≈ 4.37 rad/s

To find the maximum linear speed (v_max), we can use the relation between angular speed and linear speed for a pendulum:

v_max = ω_max * L

v_max = 4.37 rad/s * 1 m v_max ≈ 4.37 m/s

Therefore, the maximum angular speed (ω_max) of the pendulum is approximately 4.37 rad/s, and the maximum linear speed (v_max) is approximately 4.37 m/s.

c) If the length of the string is doubled, the period (T) of the pendulum changes. The period of a pendulum is inversely proportional to the square root of its length. Thus, when the length is doubled (2L), the period becomes:

T' = 2π / (√(g / (2L))) T' = √2 * (2π / ω)

Since ω remains the same, the period of the pendulum would increase by a factor of √2.

The maximum angular amplitude remains the same regardless of the length of the pendulum string.