Rates of Change — Question 6

Question 6 - Solution

Let:

• $\theta$ be the angle between the pendulum and vertical

• $x$ be the horizontal displacement of the bob from the vertical

Using trigonometry: $$x = L \sin(\theta) = 2 \sin(\theta)$$

Differentiate both sides with respect to time $t$: $$\frac{dx}{dt} = 2 \cos(\theta) \cdot \frac{d\theta}{dt}$$

Substitute: $$\theta = \frac{\pi}{6}, \quad \frac{d\theta}{dt} = 0.4 \, \text{rad/s}, \quad \cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$$

$$\frac{dx}{dt} = 2 \cdot \frac{\sqrt{3}}{2} \cdot 0.4 = \sqrt{3} \cdot 0.4 \approx 0.6928 \, \text{m/s}$$

Answer: $$\boxed{\frac{dx}{dt} \approx 0.693 \, \text{m/s}}$$