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Q53P

Expert-verifiedFound in: Page 477

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A**** ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\u200a}}{\mathbf{\text{kg}}}$ 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 t**** is in seconds. The force constant of the spring is ${\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\u200a}}{\mathit{N}}{{\mathit{m}}}^{\mathbf{-}\mathbf{1}}$****. Find **

**(a) The resonance angular frequency of the system, **

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

**(c) The amplitude of the motion.**

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

The given data can be listed below as,

- The external force equation is $F=3.00\left(\mathrm{sin}2\pi t\right)$.
- The mass of object is $m=2.00\u200a\text{kg}$.

**Amplitude**** of driven oscillator with no damping is given by:**

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

Part (c):

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

By using concept and formula from step (1), we get

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

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,

$A=\frac{\frac{\left(3\u200aN{m}^{-1}\right)}{2\u200a\text{kg}}}{\left\{{\left(6.28\u200a{\text{s}}^{-1}\right)}^{2}-{\left(3.16\u200a{\text{s}}^{-1}\right)}^{2}\right\}}\phantom{\rule{0ex}{0ex}}A=0.0509\u200a\text{m}\phantom{\rule{0ex}{0ex}}A=5.09\u200a\text{cm}$

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

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