Simple Harmonic Motion Concept Page - 11

Definition

Approximation of small displacements to SHM

Equation of SHM is given by F=−kx$F=-kx$. This is usually not exactly valid for many systems under simple harmonic oscillations. However, if the maximum displacement about the mean position is small, then the motion can be very often approximated to SHM. An example of such a scenario is simple pendulum. Restoring force for simple pendulum is given by:
F=−mgsin(θ)$F=-mg\sin (\theta )$
For small displacements, sin(θ)≈θ$\sin (\theta )\approx \theta$ and then it becomes an approximate equation of SHM.

Example

Find time period of U-Tubes by writing force equation

Example: A U-tube held vertically contains the liquid column of height L$L$. If the liquid in one of the limbs is depressed and released, what is its period of oscillations?

Solution:
Pressure =−hρg$=-h\rho g$
force = pressure×cross section area$force=pressure\times crosssectionarea$
F=−hρg×A$F=-h\rho g\times A$  ---(1)
mass of liquid column L is volume × density $massofliquidcolumnLisvolume\times density$
m=L×A×ρ$m=L\times A\times \rho$
∴F=m×a=LAρa$\therefore F=m\times a=LA\rho a$ ---(2)
From (1)and (2)
−hρgA=LAρa$-h\rho gA=LA\rho a$
a=−ghL$a={\displaystyle \frac{-gh}{L}}$
So,ω2=gL$So,{\omega }^2={\displaystyle \frac{g}{L}}$ (comparing with a=−ω2x$a=-{\omega }^{2}x$)
and T=2πω=2πLg−−√$T={\displaystyle \frac{2\pi }{\omega }}=2\pi \sqrt{{\displaystyle \frac{L}{g}}}$

Example

Time period of elastic materials using modulus of elasticity by writing force equation

Example: In the given figure, two elastic rods P$P$ and Q$Q$ are rigidly joined to end supports. A small mass m$m$ is moving with velocity v$v$ between the rods. All collisions are assumed to be elastic and the surface is given to be smooth. What will be the time period of small mass m$m$?
(A=$(A=$area of cross section, Y=$Y=$Young's modulus, L=$L=$length of each rod)$)$ ?

Solution:
KQ=YAL${\displaystyle K_Q={\displaystyle \frac{YA}{L}}}$

KP=(2Y)(A/2)L=YAL${\displaystyle K_P={\displaystyle \frac{\left(2Y\right)\left(A/2\right)}{L}}=\frac{YA}{L}}$

∴$\therefore$   KP=KQ=K=YAL${\displaystyle K_P=K_Q=K={\displaystyle \frac{YA}{L}}}$

Half oscillation is completed with P$P$ and half with Q.$Q.$ But their value of K$K$ is same. Hence we can say that in one oscillation one time period is completed with spring.
T=(2Lv)+2πmK−−−√=2Lv+2πmLAY−−−−√${\displaystyle T=\left({\displaystyle \frac{2L}{v}}\right)+2\pi \sqrt{{\displaystyle \frac{m}{K}}}={\displaystyle \frac{2L}{v}}+2\pi \sqrt{{\displaystyle \frac{mL}{AY}}}}$

Example

Find time period or frequency by writing torque equation

Example: A simple pendulum has length l$l$ and having small oscillation of angle θo${\theta }_o$ about mean position. Find the time period of this oscillation.

Solution.
Torque acting on the bob of the simple pendulum: mglsinθ$mglsin\theta$
Moment of inertia about the point of suspension: ml2$m{l}^2$
Equation of torque:
mglsinθ=−Iα$mglsin\theta =-I\alpha$  .. (1)
For small value of θ,sinθ=θ$\theta ,sin\theta =\theta$
From (1):
α∝−θ$\alpha \propto -\theta$ Since mass, g, l$l$ and Moment of Inertia is constant.
therefore w2=gl$w^2={\displaystyle \frac{g}{l}}$ 

Definition

Damping force using Stoke's approximation

Damping force in oscillations using Stoke's approximation is given by:
Fd=−bv$F_d=-bv$
where b$b$ is the damping constant and v:$v:$ speed of the body.
It acts in a direction opposite to the direction of velocity of particle and hence causes decay in the amplitude of oscillations.

Formula

Find potential energy and kinetic energy in damped oscillation as a function of time

Energy in the damped oscillation:

Differentiating the position we get the velocity

x˙(t)=−12Aeγ/2t$\dot{x}(t)=-{\displaystyle \frac{1}{2}}A{e}^{\gamma /2t}$ [γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]$[\gamma cos({\omega }_{d}t+\varphi )+2{\omega }_{d}sin({\omega }_{d}t+\varphi )]$.

Looking at the total mechanical energy (sum of the kinetic and potential energy terms), we 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$E_k={\displaystyle \frac{1}{2}}m{\dot{x}}^2$ and the potential by Ep=12kx2=12mω20x2$E_p={\displaystyle \frac{1}{2}}k{x}^2={\displaystyle \frac{1}{2}}m{\omega }_0^2{x}^2$, where we have used ω20=k/m${\omega }_0^2=k/m$.
Using the position and velocity equations are derived yields,

Ep=12mA2ω20e−γtcos2(ωdt+ϕ)$E_p={\displaystyle \frac{1}{2}}m{A}^2{\omega }_0^2{e}^{-\gamma t}co{s}^2({\omega }_{d}t+\varphi )$,

Ek=18mA2e−γt[γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]2$E_k={\displaystyle \frac{1}{8}}m{A}^2{e}^{-\gamma t}[\gamma cos({\omega }_{d}t+\varphi )+2{\omega }_{d}sin({\omega }_{d}t+\varphi ){]}^2$.

Hence the total energy is E=Ek+EP=18mA2e−γt([γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]2+4ω20cos2(ωdt+ϕ))$E=E_k+E_P={\displaystyle \frac{1}{8}}m{A}^2{e}^{-\gamma t}([\gamma cos({\omega }_{d}t+\varphi )+2{\omega }_{d}sin({\omega }_{d}t+\varphi ){]}^2+4{\omega }_0^{2}co{s}^2({\omega }_{d}t+\varphi ))$

Result

Total energy in damped oscillation and approximate for small value of damping

Energy in the damped oscillation:

Differentiating the position we get the velocity

x˙(t)=−12Aeγ/2t$\dot{x}(t)=-{\displaystyle \frac{1}{2}}A{e}^{\gamma /2t}$ [γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]$[\gamma cos({\omega }_{d}t+\varphi )+2{\omega }_{d}sin({\omega }_{d}t+\varphi )]$.

Looking at the total mechanical energy (sum of the kinetic and potential energy terms), we 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$E_k={\displaystyle \frac{1}{2}}m{\dot{x}}^2$ and the potential by Ep=12kx2=12mω20x2$E_p={\displaystyle \frac{1}{2}}k{x}^2={\displaystyle \frac{1}{2}}m{\omega }_0^2{x}^2$, where we have used ω20=k/m${\omega }_0^2=k/m$.
Using the position and velocity equations are derived yields,

Ep=12mA2ω20e−γtcos2(ωdt+ϕ)$E_p={\displaystyle \frac{1}{2}}m{A}^2{\omega }_0^2{e}^{-\gamma t}co{s}^2({\omega }_{d}t+\varphi )$,

Ek=18mA2e−γt[γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]2$E_k={\displaystyle \frac{1}{8}}m{A}^2{e}^{-\gamma t}[\gamma cos({\omega }_{d}t+\varphi )+2{\omega }_{d}sin({\omega }_{d}t+\varphi ){]}^2$.

Hence the total energy is E=Ek+EP=18mA2e−γt([γcos(ωdt+ϕ)+2ωdsin(ωdt+ϕ)]2+4ω20cos2(ωdt+ϕ))$E=E_k+E_P={\displaystyle \frac{1}{8}}m{A}^2{e}^{-\gamma t}([\gamma cos({\omega }_{d}t+\varphi )+2{\omega }_{d}sin({\omega }_{d}t+\varphi ){]}^2+4{\omega }_0^{2}co{s}^2({\omega }_{d}t+\varphi ))$

=18mA2e−γt(4ω20+2γωdsin(2ωdt+2ϕ)+γ2cos(2ωdt+2ϕ))$={\displaystyle \frac{1}{8}}m{A}^2{e}^{-\gamma t}(4{\omega }_0^2+2\gamma {\omega }_{d}sin(2{\omega }_{d}t+2\varphi )+{\gamma }^{2}cos(2{\omega }_{d}t+2\varphi ))$

=12mA2ω20e−γt[1+γ2ω0cos(2ωdt+ϕ)]$={\displaystyle \frac{1}{2}}m{A}^2{\omega }_0^2{e}^{-\gamma t}[1+{\displaystyle \frac{\gamma }{2{\omega }_0}}cos(2{\omega }_{d}t+\varphi )]$,

And ϕ=−tan−1(2ωd/γ)$\varphi =-ta{n}^{-1}(2{\omega }_d/\gamma )$.

Total energy:
E=12mA2ω20e−γt$E={\displaystyle \frac{1}{2}}m{A}^2{\omega }_0^2{e}^{-\gamma t}$

Definition

Define damping constant and find from given force or displacement equation

Damping coefficient is measure of effectiveness of damper, it reflects ability of damper to which it can resist the motion.
Damping force is given by
F=−cdxdt$F=-c{\displaystyle \frac{dx}{dt}}$

where c$c$ is the damping coefficient, given in units of newton-seconds per meter.

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