Question

Choose the correct statement (s) from the following in which k is a real, positive constant.

• A. Function f(t) = sin kt + coskt is simple harmonic having a period
• B. Function f(t) = is periodic but not simple harmonic having a period of 2s
• C. Function f(t) = is simple harmonic having a period .
• D. Function f(t) = is not periodic.

Answer 1

Answer: (A), (B), (D)

The correct options are
A Function f(t) = sin kt + coskt is simple harmonic having a period
B Function f(t) = is periodic but not simple harmonic having a period of 2s
D Function f(t) = is not periodic.

Statement (a) is correct. The function f (t) = sin kt + cos kt can be written as $f\left(t\right)=\sqrt{2}$ sin$(kt+\frac{\pi }{4})or\sqrt{2}\cos (kt-\frac{\pi }{4})$
both of which are simple harmonic. The coefficient of time t in the argument of the sine or cosine function = $\frac{2\pi }{T}$ where T is the period. Hence k =$\frac{2\pi }{T}orT=\frac{2\pi }{k}$
Statement (b) is also correct. Each term represents simple harmonic motion. The period T of term sin πt is $\pi$ = $\frac{2\pi }{T}$ or T = 2s. The period of term 2 cos 2πt is 1s, i.e., T/2 and the period of term 3 sin 3 πt is 2/3 s, i.e, T/3. The sum of two or more simple harmonic motions of different periods is not simple harmonic. The sum, however, is periodic. By the time the first term completes one cycle, the second term completes two cycles and the third term completes three cycles. Thus the sum has a period of 2s. Statement (c) is incorrect. We can write
$f\left(t\right)=coskt+2si{n}^{2}ktf\left(t\right)=coskt+(1-cos2kt)=1+coskt-cos2kt$

The period of cos kt is T = $\frac{2\pi }{T}$ and of cos 2k is $\frac{\pi }{T}$ which is T/2. As explained above, the periodof the two terms together is T = $\frac{2\pi }{T}$. The term 1 is a constant independent of time. Statement (d) is correct. Function $e^{-kt}$ decreases monotonically to zero it never becomes negative. Hence it is non-periodic.