Question

A$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{kg}}$object attached to a spring moves without friction (b=0) and is driven by an external force given by the expression$\mathit{F}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\left(sin2\pi t\right)$, where F is in Newton’s and tis in seconds. The force constant of the spring is$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{N}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$. Find

(a) The resonance angular frequency of the system,

(b) The angular frequency of the driven system, and

(c) The amplitude of the motion.

Short answer

(c) The amplitude of the motion is A = 5.09 cm.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The external force equation is $F=3.00\left(\sin 2\pi t\right)$.
• The mass of object is $m=2.00 \text{kg}$.

Significance of the amplitude

Amplitudeof driven oscillator with no damping is given by:

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

Determination of the amplitude of the motion.

Part (c):

We have to find the resonance angular frequency of the system:

Byusingconceptandformulafromstep(1),weget

$A=\frac{\frac{F_0}{m}}{\left({\omega }^2-{{\omega }_0}^2\right)}$

Referring to subparts (a) and (b) of the SID:947271-15-15.7-947271-15-53P-a and 947271-15-15.7-947271-15-53P-b

Substitute all the value in the above equation,

$$\begin{gathered} A=\frac{\frac{\left(3 N{m}^{-1}\right)}{2 \text{kg}}}{\left\{{\left(6.28 {\text{s}}^{-1}\right)}^2-{\left(3.16 {\text{s}}^{-1}\right)}^2\right\}} \\ A=0.0509 \text{m} \\ A=5.09 \text{cm} \end{gathered}$$

Hence the amplitude of the motion is A = 5.09 cm.