Question #151336
3. Under-damping linear harmonic oscillator
A block of mass m is attached to a light spring with a spring constant k and located at its equilibrium position on a smooth horizontal surface. A damping force F(v) with a magnitude linearly proportional to speed v is also acting on the block.
a) Describe your system and sketch the free-body diagram (FBD) the block immediately after it is set in motion.
b) Write the equation of motion (EOM) for the particle and its general solutions.
c) Use relevant octave script to solve provided and plot the position and velocity of the block as a function of time.
d) Describe the state of motion of the particle based on the plots in part c).
1
Expert's answer
2020-12-21T04:31:21-0500

As per the given question,

mass of block = m

Spring constant = K



Damping forceF(v)vF(v) \propto v

F(v)=γv\Rightarrow F(v) =\gamma v


md2xdt2=γv-m\frac{d^2x}{dt^2}=\gamma v


md2xdt2+γv=0\Rightarrow m\frac{d^2x}{dt^2}+\gamma v=0


d2xdt2+γvm=0\Rightarrow \frac{d^2x}{dt^2}+\frac{\gamma v}{m}=0


dvdt+γmv=0\Rightarrow \frac{dv}{dt}+\frac{\gamma}{m}v=0


Here, mγ=τ\frac{m}{\gamma}=\tau which is called relaxation time.

dvdt+1τv=0\Rightarrow \frac{dv}{dt}+\frac{1}{\tau}v=0

Taking the integration

dvv=dtτ\Rightarrow \int\frac{dv}{v}=-\int\frac{dt}{\tau}


lnv=tτ+c\Rightarrow \ln v=\frac{-t}{\tau}+c

lnv=tτ+c\Rightarrow \ln v=\frac{-t}{\tau}+c

at t =0,

C=voC=v_o

v=etτ+vo\Rightarrow v=e^{\frac{-t}{\tau}+v_o}

We know that,

v=dxdtv=\frac{dx}{dt}

dx=vdt\Rightarrow dx = v dt

dx=etτ+vodt\Rightarrow \int dx=\int e^{\frac{-t}{\tau}+v_o}dt

x=etτ+votτ+vo+C1x=\frac{e^{\frac{-t}{\tau}+v_o}}{\frac{-t}{\tau}+v_o}+C_1


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