Question

The acceleration of a certain simple harmonic oscillator is given by $a=-\left(35.28 m / s ^{2}\right) \cos 4.2 t$. 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 $\omega^{2} A=35.28$
$\omega=4.2$
$\Rightarrow A=\frac{\omega^{2} A}{(\omega)^{2}}=\frac{35.28}{(4.2)^{2}}=2.0 \,m$