Step 1: Understanding the Concept.
Instead of computing each period and then taking an LCM formula, we can find the fundamental period by asking after how many seconds each sinusoid completes a whole number of full cycles at the same time.
Step 2: Key Formula or Approach.
For a term $\sin(\omega t)$ or $\cos(\omega t)$, the ordinary frequency in Hz is $f = \frac{\omega}{2\pi}$. The signal completes exactly $k$ full cycles in a time $T$ when $f \cdot T = k$ for some positive integer $k$. The fundamental period of a sum of two sinusoids is the smallest $T$ that makes both frequencies land on a whole number of cycles at the same time.
Step 3: Detailed Explanation.
For the first term, $\omega_1 = 15\pi$, so $f_1 = \frac{15\pi}{2\pi} = 7.5$ Hz.
For the second term, $\omega_2 = 4\pi$, so $f_2 = \frac{4\pi}{2\pi} = 2$ Hz.
We need the smallest $T$ such that $f_1 T$ and $f_2 T$ are both whole numbers, that is $7.5\,T \in \mathbb{Z}$ and $2T \in \mathbb{Z}$.
Try $T = 2$ s: $f_1 T = 7.5 \times 2 = 15$, a whole number, and $f_2 T = 2 \times 2 = 4$, also a whole number. So at $T = 2$ s, the first term has completed exactly 15 cycles and the second term has completed exactly 4 cycles, both whole numbers, so the combined waveform lines back up with itself.
Now confirm nothing smaller works. Since $f_1 = 7.5 = \frac{15}{2}$ needs $T$ to be a multiple of $\frac{2}{15}$ s to give a whole number of cycles, and $f_2 = 2$ needs $T$ to be a multiple of $\frac{1}{2}$ s, the smallest $T$ satisfying both comes out to $2$ s, and no smaller common multiple of $\frac{2}{15}$ and $\frac{1}{2}$ exists, since 15 and 4 share no common factor to cancel further. So $T=2$ s is indeed the smallest, not just a common one.
Step 4: Final Answer.
The smallest time at which both sinusoidal terms simultaneously complete a whole number of cycles is 2 seconds, so the fundamental period is 2 s, matching option (D).
\[ \boxed{T = 2 \text{ s}} \]
