Respuesta :
Answer:
The number of oscillation is 36.
Explanation:
Given that
Mass = 280 g
Spring constant = 3.3 N/m
Damping constant [tex]b=8.4\times10^{-3}\ Kg/s[/tex]
We need to check the system is under-damped, critical damped and over damped by comparing b with [tex]2m\omega_{0}[/tex]
[tex]2m\omega_{0}=2m\sqrt{\dfrac{k}{m}}[/tex]
[tex]2\sqrt{km}=2\times\sqrt{3.3\times280\times10^{-3}}=1.92kg/s[/tex]
Here, b<<[tex]2m\omega_{0}[/tex]
So, the motion is under-damped and will oscillate
[tex]\omega=\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}[/tex]
The number of oscillation before the amplitude decays to [tex]\dfrac{1}{e}[/tex] of its original value
[tex]A exp(\dfrac{-b}{2m}t)=A exp(-1)[/tex]
[tex]\dfrac{b}{2m}t=1[/tex]
[tex]t = \dfrac{2m}{b}[/tex]
[tex]t=\dfrac{2\times280\times10^{-3}}{8.4\times10^{-3}}[/tex]
[tex]t=66.67\ s[/tex]
We need to calculate the time period of one oscillation
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\times3.14}{\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}}[/tex]
[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}}[/tex]
[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{3.3}{280\times10^{-3}}-\dfrac{(8.4\times10^{-3})^2}{4\times(280\times10^{-3})^2}}}[/tex]
[tex]T=1.83\ sec[/tex]
The number of oscillation is
[tex]n=\dfrac{t}{T}[/tex]
[tex]n=\dfrac{66.67}{1.83}[/tex]
[tex]n=36[/tex]
Hence, The number of oscillation is 36.
