Simple Harmonic Motion Concept Page - 3

Definition

Define epoch of SHM

Consider a particle executing simple harmonic motion along Y- axis. A simple harmonic oscillator oscillates about its mean position y=0$y=0$ to the maximum distance 'a$a$' on both the sides which is known as the amplitude of simple harmonic oscillator. The displacement at any instant t$t$ is given by y=asin(ωt+ϕ0)$y=asin(\omega t+{\varphi }_0)$. The argument of the sine function (ωt+ϕ0$\omega t+{\varphi }_0$) is called as the phase of the simple harmonic oscillator at time t$t$ and ϕ0${\varphi }_0$ is called the initial phase or epoch. ie.  at t=0,ϕ=ϕ0$t=0,\varphi ={\varphi }_0$ and this ϕ0${\varphi }_0$  is called as the epoch. ie. the displacement of an oscillating body at zero time.

Formula

Velocity in SHM as a function of time

General equation of SHM for displacement in a simple harmonic motion is:
x=Asinwt$x=Asinwt$
By definition, v=dxdt$v={\displaystyle \frac{dx}{dt}}$
or, v=Awcoswt$v=Awcoswt$

Example

Velocity as a function of displacement

General equation of SHM for displacement in a simple harmonic motion is:
x=Asinwt$x=Asinwt$
By definition, v=dxdt$v={\displaystyle \frac{dx}{dt}}$
or, v=Awcoswt$v=Awcoswt$ ... (1)

Since sin2wt+cos2wt=1$si{n}^{2}wt+co{s}^{2}wt=1$
From equation (1).

v=wA2−x2−−−−−−−√$v=w\sqrt{A^2-x^2}$

Example

Interpret direction and magnitude of velocity for different positions in SHM

Example: Two particles P$P$ and Q$Q$ describe simple harmonic motions of same period, same amplitude, along the same line about the same equilibrium position O$O$. When P$P$ and Q$Q$ are on opposite sides of O$O$ at the same distance from O$O$ they have the same speed of 1.2m/s$1.2m/s$ in the same direction, when their displacements are the same they have the same speed of 1.6m/s$1.6m/s$ in opposite directions. Find the maximum velocity in m/s$m/s$ of either particle?

Solution:
From the image,
ϕ=−θ+ωt$\varphi =-\theta +\omega t$.............................................(1)$(1)$
180−ϕ=θ+ωt$180-\varphi =\theta +\omega t$......................................(2)$(2)$
  substituting value of ϕ$\varphi$ from equn (1)$(1)$ in (2)$(2)$, we will get ωt=90$\omega t=90$. That means that particles will have to rotate by 900${90}^0$ in order to be again at same displacement.
Hence from (1),$(1),$
we get ϕ=−θ+90$\varphi =-\theta +90$ 
now from figure, 
y=Asinθ$y=A\sin \theta$.....................................................(3)$(3)$ and
y′=Asinϕ=Asin(−θ+90)=Acosθ=A1−sin2θ−−−−−−−−√$y^{\prime }=A\sin \varphi =A\sin (-\theta +90)=A\cos \theta =A\sqrt{1-{\sin }^2\theta }$
From eqn (3)$(3)$
y′=A2−y2−−−−−−√$y^{\prime }=\sqrt{A^2-y^2}$....................................................(4)$(4)$
Initially magnitude of velocity, =1.2m/s$=1.2m/s$
Therefore 1.2=ωA2−y2−−−−−−√$1.2=\omega \sqrt{A^2-y^2}$......................(5)$(5)$
and Finally 1.6=ωA2−y′2−−−−−−−√$1.6=\omega \sqrt{A^2-y^{\prime 2}}$
that gives, 1.6=ωy$1.6=\omega y$    ($($using eqn (4))$(4))$          ......................(6)$(6)$
Putting (6)$(6)$ in eqn (5)$(5)$ after simplifying,we will get ,

ωA=2m/s$\omega A=2m/s$, is the magnitude of maximum velocity which occurs

at mean position.

Shortcut

Find maximum and minimum speed in SHM from velocity as a function of time

General equation of SHM for displacement in a simple harmonic motion is:
x=Asinwt$x=Asinwt$
By definition, v=dxdt$v={\displaystyle \frac{dx}{dt}}$
or, v=Awcoswt$v=Awcoswt$ ... (1)

Clearly from equation (1) maximum velocity will be:
v=Aw$v=Aw$ for wt=0$wt=0$
and
v=0$v=0$ for wt=90$wt=90$

Which basically means velocity is maximum at the mean position and zero at the extreme position.

Result

Compare plots of velocity and displacement as a function of time

In SHM in mean position magnitude of velocity is maximum and in extreme position velocity is zero. The above graphs of velocity and displacement depicts it clearly.

Result

Angular Velocity as a Function of Time in SHM

In angular SHM equation of motion is given by:

τ=−kθ$\tau =-k\theta$
General equation for angular displacement:
θ=θosin(wt+ϕ)$\theta ={\theta }_{o}sin(wt+\varphi )$

Angular velocity =dθdt$={\displaystyle \frac{d\theta }{dt}}$
or, Angular velocity = θowcos(wt+ϕ)${\theta }_{o}wcos(wt+\varphi )$

where w=2πT$w={\displaystyle \frac{2\pi }{T}}$
Where T$T$ is time period of SHM.

Result

Acceleration as a function of displacement

Acceleration a=dvdt$a={\displaystyle \frac{dv}{dt}}$
or a=−Aw2sinwt$a=-A{w}^{2}sinwt$
or a=−w2x$a=-w^{2}x$

Example:
Acceleration-displacement graph of a particle executing SHM is as shown in the figure. What is the time period of oscillation (in sec) ?

Solution:
In SHM
a=−ω2x$a=-{\omega }^{2}x$
or
a=−Kmx$a={\displaystyle \frac{-K}{m}}x$
so, from graph
−Km=−1$-{\displaystyle \frac{K}{m}}=-1$                   (∵slope is −1)$(\because slopeis-1)$
Km=1${\displaystyle \frac{K}{m}}=1$
Time period =2πmK−−−√$=2\pi \sqrt{{\displaystyle \frac{m}{K}}}$
=2π11−−√$=2\pi \sqrt{{\displaystyle \frac{1}{1}}}$
=2π$=2\pi$

Result

Find acceleration from displacement as a function of time

In SHM for displacement has a time dependence equation in the form
x=Asinwt$x=Asinwt$

By definition, v=dxdt$v={\displaystyle \frac{dx}{dt}}$
or, v=Awcoswt$v=Awcoswt$

Acceleration is given by
a=dvdt$a={\displaystyle \frac{dv}{dt}}$
or a=−Aw2sinwt$a=-A{w}^{2}sinwt$

Diagram

Direction of acceleration in SHM

Acceleration is given by a=−ω2x$a=-{\omega }^{2}x$
Direction of acceleration is opposite to the direction of displacement.

BookMarks

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