Imagine you made a little pendulum by tying a 6.0 gram washer to a thin piece of thread and dangled it from your hand as you were given a test drive in a sports car. If the driver floored the engine and the car went from 0 to 60 mph down a straight level road in 3.2 seconds, how far off vertical would the washer hang as the car accelerated

The displacement of the pendulum from equilibrium position is 0.855 of the length of the pendulum.

Explanation:

A pendulum's motion can be modelled as a simple harmonic motion and the differential equation describing simple harmonic motion is given as

(d²x/dt²) + (k/m)x = 0

(d²x/dt²) = acceleration of the pendulum

x = displacement of the pendulum from equilibrium position

k = spring constant of the pendulum = (mg/L) = (0.006×9.8/L) = (0.0588/L)

L = length of the pendulum

m = mass of the pendulum = 6 g = 0.006 kg

(k/m) = (g/L) = (9.8/L)

To calculate the pendulum's acceleration

Initial velocity of the car = 0 mph = 0 m/s

Final velocity of the car = 60 mph = 26.822 m/s

Time taken for velocity change = 3.2 s

Acceleration = (Δv/Δt) = (26.822/3.2) = 8.382 m/s²

(d²x/dt²) + (k/m)x = 0

Becomes

8.383 + (9.8/L)x = 0

(9.8x/L) = -8.382

9.8x = -8.382L

x = -(8.382/9.8) = -0.855 L (the minus sign shows the direction of movement of the pendulum)

Ignoring the sign and only focusing on the magnitude of the pendulum's displacement from equilibrium position, the displacement of the pendulum from equilibrium position is 0.855 of the length of the pendulum.

Hope this Helps!!!