Question

A mass - spring system undergoes an oscillation with an amplitude A = 20 cm and a speed at the equilibrium position of 4 m / s . What is the maximum acceleration of the system ?

Answer 1

ANSWER:

5th option: 80 m/s²

STEP-BY-STEP EXPLANATION:

Given:

Amplitude (A) = 20 cm = 0.2 m

speed in the equilibrium position (v) = 4m/s

To calculate the acceleration, the first thing is to calculate the frequency, using the following formula (maximum speed, i.e. the speed in the equilibrium position):

$\begin{gathered} v=2\pi fA \\ \\ \text{ We replacing} \\ \\ 4=(2)(3.14)(0.2)f \\ \\ f=\frac{4}{(2)(3.14)(0.2)} \\ \\ f=3.185\text{ Hz} \end{gathered}$

Now, we calculate the value of the maximum acceleration using the following formula:

$\begin{gathered} a=(2\pi f)^2A \\ \\ \text{ We replacing} \\ \\ a=0.2\cdot((2)(3.14)(3.185))^2 \\ \\ a=80.01\cong80\text{ m/s}^2 \end{gathered}$

Therefore, the maximum acceleration of the system is equal to 80 m/s² so the correct answer is 5th option: 80 m/s²