Question

Damping is negligible for a 0.150 kgobject hanging from a light, $\mathbf{6}\mathbf{.}\mathbf{30}\mathbf{}\mathit{N}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$spring. A sinusoidal force with amplitude of 1.70 Ndrives the system. At what frequency will the force make the object vibrate with amplitude of 0.440 m?

Short answer

The resonance frequency is either$f=1.31 \text{Hz}$or $f=0.641 \text{Hz}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The force constant is $K=6.30 N{m}^{-1}$.
• The mass of object is$m=0.150 \text{kg}$ .
• The external force is, $F_0=17.0 \text{N}$.
• The amplitude of sinusoidal wave is $A=0.440 \text{m}$.

Significance of the amplitude

The equation for the amplitude of forced oscillations is expressed as,

$\mathit{A}\mathbf{=}\frac{\frac{{\mathbf{F}}_{\mathbf{0}}}{\mathbf{m}}}{\sqrt{{\left({\omega }^2-{{\omega }_0}^2\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{b\omega }{m}\right)}^{\mathbf{2}}}}$

For b = 0,

Amplitudeof driven oscillator with no damping is given by:

role="math" localid="1663674316174" $$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{\pm }\frac{\frac{{\mathbf{F}}_{\mathbf{0}}}{\mathbf{m}}}{\left({\omega }^2-{{\omega }_0}^2\right)} \\ {\mathit{\omega }}^{\mathbf{2}}\mathbf{=}{{\mathbf{\omega }}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{\pm }\frac{\frac{{\mathbf{F}}_{\mathbf{0}}}{\mathbf{m}}}{\mathbf{A}} \\ {\mathit{\omega }}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{K}}{\mathbf{m}}\mathbf{\pm }\frac{{\mathbf{F}}_{\mathbf{0}}}{\mathbf{m}\mathbf{A}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

…(1)

Determination of the frequency

Substitute all the value in the equation (1),

$$\begin{gathered} {\omega }^2=\frac{K}{m}\pm \frac{F_0}{mA} \\ {\omega }^2=\frac{6.30 {\text{Nm}}^{-1}}{0.150 \text{kg}}\pm \frac{1.70 \text{N}}{0.150 \text{kg}\times 0.440 \text{m}} \end{gathered}$$

This yield: $\omega =8.23 \text{rad}{\text{s}}^{-1}$or $\omega =4.03 rad{s}^{-1}$.

Then:

$f=\frac{\omega }{2\pi }$

For$\omega =8.23 rad{s}^{-1}$

$$\begin{gathered} f=\frac{8.23 \text{rad}{\text{s}}^{-1}}{2\pi } \\ f=1.31 \text{Hz} \end{gathered}$$

For $\omega =4.03 \text{rad}{\text{s}}^{-1}$

$$\begin{gathered} f=\frac{4.03 rad{s}^{-1}}{2\pi } \\ f=0.641 \text{Hz} \end{gathered}$$

$f=1.31 \text{Hzor}f=0.641 \text{Hz}$