Question

A vibratory motion is represented by $\mathrm{x}=2\mathrm{Acos\omega t}+\mathrm{Acos}\left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)+\mathrm{Acos}(\mathrm{\omega t}+\mathrm{\pi })+\left(\frac{\mathrm{A}}{2}\right)\cos \left(\mathrm{\omega t}+\frac{3\mathrm{\pi }}{2}\right)$. The resultant amplitude of the motion is

• A. $\frac{9\mathrm{A}}{2}$
• B. $\frac{\sqrt{5}\mathrm{A}}{2}$
• C. $\frac{5\mathrm{A}}{2}$
• D. $2\mathrm{A}$

Answer 1

The correct option is B

$\frac{\sqrt{5}\mathrm{A}}{2}$

Step 1: Given data

Vibratory motion, $\mathrm{x}=2\mathrm{Acos\omega t}+\mathrm{Acos}\left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)+\mathrm{Acos}(\mathrm{\omega t}+\mathrm{\pi })+\left(\frac{\mathrm{A}}{2}\right)\cos \left(\mathrm{\omega t}+\frac{3\mathrm{\pi }}{2}\right)$

Step 2: To find

We have to determine the resultant amplitude of the motion

Step 3: Calculation

Vibrational motion occurs whenever a body moves back and forth about its mean position. Vibratory motion is defined as any item moving/swinging back and forth, vertically and horizontally, pulsing, and so on.

$\therefore \cos (\mathrm{A}+\mathrm{B})=\mathrm{cosA}\mathrm{cosB}-\mathrm{sinA}\sin \mathrm{B}$

$$\begin{gathered} \Rightarrow \mathrm{x}=2\mathrm{Acos\omega t}+\mathrm{Acos}\left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)+\mathrm{Acos}(\mathrm{\omega t}+\mathrm{\pi })+\left(\frac{\mathrm{A}}{2}\right)\cos \left(\mathrm{\omega t}+\frac{3\mathrm{\pi }}{2}\right) \\ =2\mathrm{Acos\omega t}–\mathrm{Asin\omega t}–\mathrm{Acos\omega t}+\left(\frac{\mathrm{A}}{2}\right)\mathrm{sin\omega t} \\ =\mathrm{Acos\omega t}-\left(\frac{\mathrm{A}}{2}\right)\mathrm{sin\omega t} \end{gathered}$$

Now,

The resultant amplitude be:

$$\begin{gathered} \Rightarrow \mathrm{A}’=\surd ({\mathrm{A}}^2+{(-\mathrm{A}/2)}^2) \\ =\left(\frac{\sqrt{5}}{2}\right)\mathrm{A} \\ =\frac{\sqrt{5}\mathrm{A}}{2} \end{gathered}$$

Therefore, option B is the correct choice.