The problem involves a weight attached to a spring oscillating in simple harmonic motion. The displacement of the spring is given by the equation:

$x = 14 \cos(t)$

where $x$ is the displacement in centimeters, and $t$ is the time in seconds.

Part (a): Displacement when $t = \frac{\pi}{3}$

Given: $t = \frac{\pi}{3}$

Substituting $t$ into the displacement equation:

$x = 14 \cos\left(\frac{\pi}{3}\right)$

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$:

$x = 14 \times \frac{1}{2} = 7 \, \text{cm}$

Part (b): Displacement when $t = \frac{3\pi}{4}$

Given: $t = \frac{3\pi}{4}$

Substituting $t$ into the displacement equation:

$x = 14 \cos\left(\frac{3\pi}{4}\right)$

Since $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$:

$x = 14 \times \left(-\frac{\sqrt{2}}{2}\right) = -14 \times 0.7071 \approx -9.9 \, \text{cm}$

So, the displacement when $t = \frac{3\pi}{4}$ is approximately:

$\boxed{-9.9 \, \text{cm}}$

Would you like further details or have any questions? Here are some related questions for further exploration:

1. How do you determine the amplitude of an oscillating spring?
2. What does the negative sign in displacement indicate in terms of the spring's movement?
3. How do phase shifts affect the motion of a spring?
4. What is the period of the spring's motion in this scenario?
5. How would the displacement change if the spring constant were different?

Tip: For trigonometric evaluations, always check the angle's quadrant to determine the correct sign of the trigonometric function value.