Question

an object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s². find the amplitude.

a. 28 m
b. 5 m
c. 36 m
d. 62 m
e. 100 m

Explanation:

Step1: Recall formulas for SHM

The maximum speed in simple - harmonic motion is $v_{max}=\omega A$ and the maximum acceleration is $a_{max}=\omega^{2}A$.

Step2: Express $\omega$ from the speed formula

From $v_{max}=\omega A$, we can get $\omega=\frac{v_{max}}{A}$.

Step3: Substitute $\omega$ into the acceleration formula

Substitute $\omega=\frac{v_{max}}{A}$ into $a_{max}=\omega^{2}A$. We have $a_{max}=(\frac{v_{max}}{A})^{2}A=\frac{v_{max}^{2}}{A}$.

Step4: Solve for amplitude $A$

Rearrange the formula $a_{max}=\frac{v_{max}^{2}}{A}$ to get $A = \frac{v_{max}^{2}}{a_{max}}$.
Given $v_{max}=4.3\ m/s$ and $a_{max}=0.65\ m/s^{2}$, then $A=\frac{4.3^{2}}{0.65}=\frac{18.49}{0.65}\approx28\ m$.

Answer:

A. 28 m