Question

Which of the following functions has period $2\pi$

• A. $y = \sin\left(2\pi t + \frac{\pi}{3}\right) + 2 \sin\left(3\pi t + \frac{\pi}{4}\right) + 3 \sin 5\pi t$
• B. $y = \sin \frac{\pi}{3}t + \sin \frac{\pi}{4}t$
• C. $y = \sin t + \cos 2t$
• D. None of these

Answer 1

Answer: (C)

(c)

Answer 2

The period of a function $f(t)$ is the smallest positive value $T$ such that $f(t+T) = f(t)$ for all $t$ in the domain of $f$. For a sum of periodic functions, $y(t) = \sum_{i=1}^n f_i(t)$, where each $f_i(t)$ has a period $T_i$, the period of $y(t)$ is the least common multiple (LCM) of the individual periods $T_1, T_2, \dots, T_n$, provided that the ratios of the periods are rational. If the ratio of any two periods is irrational, the sum is generally not periodic.

Let's analyze each option:

(a) $y = \sin\left(2\pi t + \frac{\pi}{3}\right) + 2 \sin\left(3\pi t + \frac{\pi}{4}\right) + 3 \sin 5\pi t$

The periods of the terms are: $T_1 = \frac{2\pi}{|2\pi|} = 1$ $T_2 = \frac{2\pi}{|3\pi|} = \frac{2}{3}$ $T_3 = \frac{2\pi}{|5\pi|} = \frac{2}{5}$

The periods are 1, 2/3, and 2/5. These are rational numbers, so their ratios are rational. The period of $y$ is LCM$(1, 2/3, 2/5)$. LCM of rational numbers $\frac{p_1}{q_1}, \frac{p_2}{q_2}, \dots, \frac{p_n}{q_n}$ is $\frac{\text{LCM}(p_1, p_2, \dots, p_n)}{\text{HCF}(q_1, q_2, \dots, q_n)}$. LCM$(1, 2/3, 2/5) = \text{LCM}(1/1, 2/3, 2/5) = \frac{\text{LCM}(1, 2, 2)}{\text{HCF}(1, 3, 5)} = \frac{2}{1} = 2$. The period of function (a) is 2.

(b) $y = \sin \frac{\pi}{3}t + \sin \frac{\pi}{4}t$

The periods of the terms are: $T_1 = \frac{2\pi}{|\pi/3|} = \frac{2\pi}{\pi/3} = 6$ $T_2 = \frac{2\pi}{|\pi/4|} = \frac{2\pi}{\pi/4} = 8$

The periods are 6 and 8. These are rational numbers, so their ratio is rational. The period of $y$ is LCM$(6, 8) = 24$. The period of function (b) is 24.

(c) $y = \sin t + \cos 2t$

The periods of the terms are: $T_1 = \frac{2\pi}{|1|} = 2\pi$ $T_2 = \frac{2\pi}{|2|} = \pi$

The periods are $2\pi$ and $\pi$. The ratio $T_1/T_2 = 2\pi/\pi = 2$, which is rational. The period of $y$ is LCM$(2\pi, \pi)$. We are looking for the smallest positive value $T$ that is a multiple of both $2\pi$ and $\pi$. Let $T = n \cdot 2\pi$ for some positive integer $n$. Let $T = m \cdot \pi$ for some positive integer $m$. So, $n \cdot 2\pi = m \cdot \pi$, which implies $2n = m$. To find the smallest positive $T$, we need the smallest positive integers $n$ and $m$ satisfying $2n=m$. The smallest positive integer for $n$ is 1, which gives $m=2$. With $n=1$, $T = 1 \cdot 2\pi = 2\pi$. With $m=2$, $T = 2 \cdot \pi = 2\pi$. The smallest common multiple is $2\pi$. The period of function (c) is $2\pi$.

(d) None of these.

Since function (c) has a period of $2\pi$, option (d) is incorrect.

Therefore, the function $y = \sin t + \cos 2t$ has a period of $2\pi$.