Question

Given \(x(t) = 4\sin(15\pi t) + 7\cos(4\pi t)\), where \(t\) is in seconds.
The fundamental period of \(x(t)\) is _____ seconds.

• A. 15
• B. 60
• C. 30
• D. 2

Hint:

Find the individual period of each sinusoidal term, then take the least common multiple of the two periods.

Answer 1

The Correct Option is D

Step 1: Understanding the Question.
We are given \(x(t) = 4\sin(15\pi t) + 7\cos(4\pi t)\), a sum of two sinusoids at different frequencies, and need the fundamental period of the combined signal, that is the smallest \(T\) such that \(x(t+T) = x(t)\) for every \(t\).

Step 2: Key Formula or Approach.
Each sinusoid \(\sin(\omega t)\) or \(\cos(\omega t)\) has its own period \(T = \frac{2\pi}{\omega}\). When two periodic signals with periods \(T_1\) and \(T_2\) are added, the sum is periodic only if \(T_1\) and \(T_2\) are commensurate (their ratio is a rational number), and in that case the fundamental period of the sum is the least common multiple (LCM) of \(T_1\) and \(T_2\), the smallest time after which both sinusoids simultaneously complete a whole number of their own cycles.

Step 3: Detailed Explanation.
For the first term, \(4\sin(15\pi t)\), the angular frequency is \(\omega_1 = 15\pi\), so its period is
\[ T_1 = \frac{2\pi}{15\pi} = \frac{2}{15} \text{ s} \]
For the second term, \(7\cos(4\pi t)\), the angular frequency is \(\omega_2 = 4\pi\), so its period is
\[ T_2 = \frac{2\pi}{4\pi} = \frac{1}{2} \text{ s} \]
Both periods are rational numbers of seconds, so their ratio is rational and the sum is indeed periodic. To find the LCM of \(\frac{2}{15}\) and \(\frac{1}{2}\), write it as the smallest \(T\) that is an integer multiple of both. If \(T = m T_1 = n T_2\) for positive integers \(m, n\) with no common factor, then
\[ \frac{m}{n} = \frac{T_2}{T_1} = \frac{1/2}{2/15} = \frac{15}{4} \]
Since \(15\) and \(4\) share no common factor, the smallest such integers are \(m = 15\), \(n = 4\). Substituting back,
\[ T = n T_2 = 4 \times \frac{1}{2} = 2 \text{ s} \]
As a check, \(T = m T_1 = 15 \times \frac{2}{15} = 2\) s as well, so both give the same answer, confirming \(T = 2\) s.

Step 4: Final Answer.
Options (A) 15 s, (B) 60 s, and (C) 30 s do not match the LCM computed above; they come from mixing up the two individual periods or angular frequencies without correctly taking the LCM of the two periods. The correct fundamental period, the smallest time after which both the \(15\pi\) term and the \(4\pi\) term simultaneously repeat, is 2 seconds.
\[ \boxed{T = 2 \text{ s}} \]