Step 1: Understanding the Concept:
The given equation is a superposition of two simple harmonic motions of the same frequency and along the same line. The resultant motion is also a simple harmonic motion. The amplitude of this resultant motion can be found by combining the two components.
Step 2: Key Formula or Approach:
An expression of the form $x = a \sin(\omega t) + b \cos(\omega t)$ can be written in the form of a single SHM equation $x = A \sin(\omega t + \phi)$, where the resultant amplitude $A$ is given by:
\[ A = \sqrt{a^2 + b^2} \]
The phase angle $\phi$ is given by $\tan \phi = \frac{b}{a}$.
Step 3: Detailed Explanation:
The given displacement equation is:
\[ x = 3 \sin(2t) + 4 \cos(2t) \]
This equation is of the form $x = a \sin(\omega t) + b \cos(\omega t)$.
By comparing the equations, we have:
- $a = 3$
- $b = 4$
- The angular frequency is $\omega = 2$.
The amplitude of the resultant SHM is given by the formula:
\[ A = \sqrt{a^2 + b^2} \]
Substitute the values of a and b:
\[ A = \sqrt{3^2 + 4^2} \]
\[ A = \sqrt{9 + 16} \]
\[ A = \sqrt{25} \]
\[ A = 5 \]
The units of amplitude will be the same as the units of x.
Step 4: Final Answer:
The amplitude of the particle is 5. Therefore, option (D) is correct.