Question

A simple harmonic motion is represented by $x=7sin\omega t+$24 $co{s}^2\frac{\omega t}{2}$, then the amplitude of motion is approximately

• A. 24
• B. 31
• C. 14
• D. 25

Answer 1

The correct option is C 14

Given, $x=7sin\omega t+$24 $co{s}^2\frac{\omega t}{2}$
$x=7sin\left(\omega t\right)+24\left\{\frac{1+cos\omega t}{2}\right\}$
$=12+7sin\left(\omega t\right)+12cos\left(\omega t\right)$
Amplitude $=\sqrt{A^2+B^2}$
$A=\sqrt{7^2+{12}^2}=\sqrt{193}=13.89\approx 14$