Question

The acceleration of a certain simple harmonic oscillator is given by $a=-\left(35.28m/s^2\right)\cos 4.2t$. The amplitude of the simple harmonic motion is :

• A. 2.0 m
• B. 8.4 m
• C. 16.8 m
• D. 17.64 m

Answer 1

Answer: (A)

Given ${\omega }^{2}A=35.28$
$\omega =4.2$
$\Rightarrow A=\frac{{\omega }^{2}A}{(\omega {)}^2}=\frac{35.28}{(4.2{)}^2}=2.0m$