Question

A vertical mass-spring oscillator has an amplitude of $\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{}\mathit{m}$ and a period of $\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathit{s}$. (a) What is the maximum speed of the mass? (b) What is the maximum acceleration of the mass?

Short answer

a) The maximum speed of the mass is$0.94m/s$.

b) The maximum acceleration of the mass is$14.8m/s^2$.

Step-by-step solution

Identification of given data

Given data can be listed below,

• Amplitude, $A=0.06m$

• Period, $T=0.4s$

Evaluating angular frequency of the mass-spring system

Expression for the period of oscillation is given as,

$$\begin{gathered} T=\frac{2\pi }{\omega } \\ \omega =\frac{2\pi }{T} \end{gathered}$$

Substituting $0.4s$for $T$ in the above equation

$$\begin{gathered} \omega =\frac{2\pi }{0.4} \\ =15.7rad/s \end{gathered}$$

Evaluating maximum speed of mass

Expression for the maximum speed of the spring-mass system is

${\omega }_{max}=A$

Substituting $0.06m$for $A$, $15.7rad/s$for $\omega$in the above equation,

$$\begin{gathered} v_{max}=0.06\times 15.7 \\ =0.94m/s \end{gathered}$$

Evaluating maximum acceleration of mass

Expression for the maximum acceleration of the spring-mass system is,

${\omega }_{max}=A{\omega }^2$

Substituting $0.06m$for $A$, $15.7rad/s$for $\omega$in the above equation,

$$\begin{gathered} {\omega }_{max}=0.06\times 15.7^2 \\ =14.8m/s^2 \end{gathered}$$