Respuesta :
Answer:
[tex] KE=0.7341\ J[/tex]
Explanation:
Given:
- spring constant, [tex]k=350\ N.m^{-1}[/tex]
- mass of the block attached, [tex]m=0.24\ kg[/tex]
- damping constant, [tex]b=0.41\ kg.s^{-1}[/tex]
- amplitude of oscillation, [tex]A=0.075\ m[/tex]
Now the frequency of damped oscillation is given as:
[tex]\omega'=\sqrt{\frac{k}{m} -\frac{b^2}{4m^2} }[/tex]
[tex]\omega'=\sqrt{\frac{350}{0.24} -\frac{0.41^2}{4\times 0.24^2} }[/tex]
[tex]\omega'=38.1786\ rad.s^{-1}[/tex]
Now time period of one oscillation:
[tex]T=\frac{2\pi}{\omega'}[/tex]
[tex]T=\frac{2\pi}{38.1786}[/tex]
[tex]T=0.1646\ s[/tex]
We know the equation of motion for the damped harmonic motion of a mass attached to a spring is given as:
[tex]x=A.e^{-(b.t/2m)}.cos\ \omega'.t[/tex] ........................(1)
From above we find the position of the mass:
[tex]x=0.075\times e^{-(0.41\times 0.1646)/(2\times 0.24)}\times cos\ (38.1786\times 0.1646)[/tex]
[tex]x=0.06477\ m[/tex] is the position after 1 cycle.
So, the Kinetic energy can be given as:
[tex]KE=\frac{1}{2} k.x^2[/tex]
[tex]KE=0.5\times 350\times 0.06477^2[/tex]
[tex] KE=0.7341\ J[/tex]
