Simple Harmonic Motion Concept Page - 2

Definition

Fourier's statement

Fourier's statement: Any periodic function can be represented in a linear combination of sinusoidal functions.

Definition

Simple Harmonic Motion

Definition

Time-period from equation of SHM

Equation of SHM: F=−mω2x$F=-m{\omega }^{2}x$
Angular frequency =ω$=\omega$
Frequency f=ω2π$f={\displaystyle \frac{\omega }{2\pi }}$
Time-period T=2πω$T={\displaystyle \frac{2\pi }{\omega }}$
Example:
Acceleration-displacement graph of a particle executing SHM is as shown in the figure. Find the time period of oscillation.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$=2\pi$

Definition

Applications of SHM

Some applications of SHM are:

• Simple harmonic motion of a pendulum is used for the measurement of time.
• Tuning of the musical instrument is done with the vibrating tuning fork which executes simple harmonic motion.
• Wave is a consequence of simple harmonic motion. Study of waves is indirectly the study of simple harmonic motion.
• Molecules are in simple harmonic motion. This study is called vibration spectroscopy.

Definition

Characteristics of SHM

1. A restoring force must act on the body.
2. Body must have acceleration in a direction opposite to the displacement and the acceleration must be directly proportional to displacement.
3. The system must have inertia (mass).
4. SHM is a type of oscillatory motion.
5. It is a particular case of preodic motion.
6. It can be represented by a simple sine or cosine function

Example

Restoring force as a function of time

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path.

Restoring force F=kx$F=kx$, is such kind of force and is conservative.

F=−kAsinwt$F=-kAsinwt$

Example

Extreme Position(Amplitude) of a particle performing SHM

Example : A particle moves along y-axis according to the equation y (in cm)=3sin100πt+8sin250πt−6$=3\sin 100\pi t+8{\sin }^{2}50\pi t-6$.Find whether the motion is simple harmonic or not.Also, calculate amplitude of particle and its mean position.Solution:
The given equation can be written as
y=3sin100πt+(4−4cos100πt)−6$y=3\sin 100\pi t+(4-4\cos 100\pi t)-6$
   =3sin100πt+4sin(100πt+π/2)−2$=3\sin 100\pi t+4\sin (100\pi t+\pi /2)-2$
or y=5sin(100π−53∘)−2$y=5\sin (100\pi -{53}^{\circ })-2$
ymax=5−2=3$y_{max}=5-2=3$
ymin=−5−2=−7$y_{min}=-5-2=-7$
Mean position=ymax+ymin2=−2${\displaystyle ={\displaystyle \frac{y_{max}+y_{min}}{2}}=-2}$ cm

Example

Use phase to understand relative position between two SHMs

Example: What is the minimum phase difference between two SHMs y1=sin(π/6)sin(ωt)+sin(π/3)cos(ωt)$y_1=sin(\pi /6)sin(\omega t)+sin(\pi /3)cos(\omega t)$ ; y2=cos(π/6)sin(ωt)+cos(π/3)cos(ωt)$y_2=cos(\pi /6)sin(\omega t)+cos(\pi /3)cos(\omega t)$ ?

Solution:
y1=sin(ωt)sin(π6)+cos$y_1=sin(\omega t)sin({\displaystyle \frac{\pi }{6}})+cos$ ωtsinπ3$\omega tsin{\displaystyle \frac{\pi }{3}}$
=sin$=sin$ ωtcos(π3)+cos$\omega tcos({\displaystyle \frac{\pi }{3}})+cos$ ωtsinπ3$\omega tsin{\displaystyle \frac{\pi }{3}}$
=sin(ωt+π3)$=sin(\omega t+{\displaystyle \frac{\pi }{3}})$
y2=sin(ωt+π6)$y_2=sin(\omega t+{\displaystyle \frac{\pi }{6}})$
Thus phase difference is 

π3−π6=π6${\displaystyle \frac{\pi }{3}}-{\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\pi }{6}}$

Shortcut

Find time taken to travel between two given positions in SHM

Example: A particle executes SHM along a straight line with mean position at x=0$x=0$ and with a period of 20$20$ sec and amplitude of 5$5$ cm. Find the shortest time taken by it to go from x=4$x=4$cm to x=−3$x=-3$cm ?

Solution:
x=A$x=A$sin( ωt+ϕ$\omega t+\varphi$)
ω=2π20=2π20=π10radsec$\omega ={\displaystyle \frac{2\pi }{20}}={\displaystyle \frac{2\pi }{20}}={\displaystyle \frac{\pi }{10}}{\displaystyle \frac{rad}{sec}}$
let at t=0,x=4$t=0,x=4$, thus
4=5$4=5$ sin ϕ$\varphi$
sin−145=ϕ$si{n}^{-1}{\displaystyle \frac{4}{5}}=\varphi$
Now for at t=t1$t=t_1$, let x=−3$x=-3$, thus we have
−35=sin(ωt1+ϕ)${\displaystyle \frac{-3}{5}}=sin(\omega t_1+\varphi )$
or
sin−1(−35)−sin−1(45)=ωt1$si{n}^{-1}({\displaystyle \frac{-3}{5}})-si{n}^{-1}({\displaystyle \frac{4}{5}})=\omega t_1$
10(−0.643−0.927)π=t1$10{\displaystyle \frac{(-0.643-0.927)}{\pi }}=t_1$
Solving we get t1=5 sec$t_1=5sec$

Diagram

Shift of displacement-time plot with change in phase

In the given plot, phase difference is π/4$\pi /4$
x1(t)=Asin(ωt)$x_1(t)=A\sin (\omega t)$
x2(t)=Asin(ωt+π/4)$x_2(t)=A\sin (\omega t+\pi /4)$

Law

Displacement as a function of time is a simple harmonic motion

Standard equation of simple harmonic motion is:
a=−w2x$a=-w^{2}x$
Any general equation satisfying the above criterion represents a simple harmonic motion.
i.e. x=Asinwt$x=Asinwt$

Formula

Angular displacement as a function of time

In angular SHM equation of motion is given by:

τ=−kθ$\tau =-k\theta$
General equation for angular displacement:
θ=θosin(wt+ϕ)

BookMarks

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