Derive the expression of the mechanical energy for under damped system.
Answers
Explanation:
1 Introduction
In another node (damped-harmonic-oscillator) we derived the motion of an under-damped harmonic oscillator and found
x(t)=Ae−γ/2tcos(ωdt+ϕ),
where ωd=ω02−γ2/4, γ is the damping rate, and ω0 is the angular frequency of the oscillator without damping. Here we will investigate the energy of the system. Clearly this is no longer a closed system so we should expect the energy to dissipate to the environment and the motion to cease eventually.
2 Energy in the underdamped oscillator
Differentiating the position we get the velocity
x˙(t)=−12Ae−γ/2t[γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)].
Looking at the total mechanical energy (sum of the kinetic and potential energy terms), we'd expect this decay away with time as the velocity dependent damping is removing energy from the mechanical system. The kinetic energy is given as usual by Ek=12mx˙2 and the potential by Ep=12kx2=12mω02x2, where we have used ω02=k/m. Using the position and velocity equations we derived yields
Ep=12mA2ω02e−γtcos2(ωdt+ϕ),Ek=18mA2e−γt[γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]2.
Here is ur✔ans..
In another node (damped-harmonic-oscillator) we derived the motion of an under-damped harmonic oscillator and found
x(t)=Ae−γ/2tcos(ωdt+ϕ),
where ωd=ω02−γ2/4, γ is the damping rate, and ω0 is the angular frequency of the oscillator without damping. Here we will investigate the energy of the system. Clearly this is no longer a closed system so we should expect the energy to dissipate to the environment and the motion to cease eventually.
2 Energy in the underdamped oscillator
Differentiating the position we get the velocity
x˙(t)=−12Ae−γ/2t[γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)].
Looking at the total mechanical energy (sum of the kinetic and potential energy terms), we'd expect this decay away with time as the velocity dependent damping is removing energy from the mechanical system. The kinetic energy is given as usual by Ek=12mx˙2 and the potential by Ep=12kx2=12mω02x2, where we have used ω02=k/m. Using the position and velocity equations we derived yields
Ep=12mA2ω02e−γtcos2(ωdt+ϕ),Ek=18mA2e−γt[γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]2.
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