Question

Suppose a pendulum of length L meters makes an angle of θ radians with the vertical, as in the figure. It can be shown that as a function of time, θ satisfies the differential equation d^2θ/dt^2 + g/L sin θ = 0, where g = 9.8 m/s^2 is the acceleration due to gravity. For θ near zero we can use the linear approximation sin (θ) is approximately equal to θ to get a linear differential equation d^2θ/dt^2 + g/L θ = 0.Use the linear differential equation to answer the following questions.

a. Determine the equation of motion for a pendulum of length 1.5 meters having initial angle 0.4 radians and initial angular velocity dθ/dt = 0.2 radians per second. θ(t) = radians
b. What is the period of the pendulum? That is, what is the time for one swing back and forth? Period = seconds

Answer 1

The equation of motion for the pendulum is θ(t) = 0.4 cos(√(9.8/1.5)t) + 0.2 sin(√(9.8/1.5)t). The period of the pendulum is approximately 2.65 seconds.

Step-by-step explanation:

To determine the equation of motion for a pendulum of length 1.5 meters with an initial angle of 0.4 radians and initial angular velocity of 0.2 radians per second, we can use the linear differential equation d^2θ/dt^2 + (g/L)θ = 0. Plugging in the values, we get d^2θ/dt^2 + (9.8/1.5)θ = 0. To solve this equation, we assume a solution of the form θ(t) = A sin(ωt+ϕ), where A is the amplitude, ω is the angular frequency, and ϕ is the phase constant.

By comparing the coefficients, we can determine the equation of motion to be θ(t) = 0.4 cos(√(9.8/1.5)t) + 0.2 sin(√(9.8/1.5)t).

The period of the pendulum can be determined using the formula T = 2π/ω, where ω = √(g/L). Plugging in the values, we get T = 2π/√(9.8/1.5), which is approximately 2.65 seconds.