what is the physical position at z=0? b. what is the physical position corresponding to r=0

What is Simple Harmonic Motion?

Elementary Harmonic Movement or SHM is divers as a motion in which the restoring force is directly proportional to the displacement of the torso from its hateful position. The direction of this restoring forcefulness is ever towards the hateful position. The acceleration of a particle executing elementary harmonic movement is given past, a(t) = -ω²x(t). Here, ω is the athwart velocity of the particle.

Table of Contents

• Deviation Between Elementary Harmonic, Periodic and Oscillation Move
• Types of Elementary Harmonic Movement
• General Terms
• Differential Equation
• Athwart SHM
• Quantitative Analysis
• Necessary weather condition
• Free energy in SHM
• Geometrical Interpretations
• Horizontal Phasor

Simple Harmonic, Periodic and Oscillation Move

Elementary harmonic motility tin be described as an oscillatory motion in which the acceleration of the particle at any position is directly proportional to the displacement from the mean position. Information technology is a special case of oscillatory motility.

All the Unproblematic Harmonic Motions are oscillatory and also periodic but not all oscillatory motions are SHM. Oscillatory motion is also called the harmonic move of all the oscillatory motions wherein the most of import i is simple harmonic motion (SHM).

In this type of oscillatory motion displacement, velocity and acceleration and force vary (w.r.t fourth dimension) in a style that can be described by either sine (or) the cosine functions collectively called sinusoids.

Likewise Read:

• Simple Pendulum Concepts
• Leap-Mass Arrangement

The study of Uncomplicated Harmonic Motion is very useful and forms an important tool in understanding the characteristics of audio waves, light waves and alternating currents. Any oscillatory motion which is not simple Harmonic tin can be expressed as a superposition of several harmonic motions of different frequencies.

Departure between Periodic, Oscillation and Simple Harmonic Motion

Periodic Motion

• A movement repeats itself after an equal interval of time. For instance, uniform circular motion.
• At that place is no equilibrium position.
• There is no restoring force.
• In that location is no stable equilibrium position.

Oscillation Movement

• To and fro motility of a particle about a mean position is called an oscillatory motility in which a particle moves on either side of equilibrium (or) hateful position is an oscillatory motion.
• It is a kind of periodic motility bounded between two extreme points. For example, Oscillation of Simple Pendulum, Spring-Mass Organisation.
• The object will keep on moving betwixt two farthermost points nigh a stock-still point is called mean position (or) equilibrium position along whatever path. (the path is not a constraint).
• At that place will exist a restoring force directed towards equilibrium position (or) hateful position.
• In an oscillatory motion, the net forcefulness on the particle is nil at the hateful position.
• The hateful position is a stable equilibrium position.

Unproblematic Harmonic Motion or SHM

• It is a special instance of oscillation along with straight line betwixt the two extreme points (the path of SHM is a constraint).
• Path of the object needs to exist a straight line.
• There will be a restoring force directed towards equilibrium position (or) mean position.
• Mean position in Uncomplicated harmonic motion is a stable equilibrium.

Atmospheric condition for SHM:

\(\begin{array}{l}\brainstorm{matrix} \overrightarrow{F}\propto -\overrightarrow{10} \\ \overrightarrow{a}\,\,\propto -\overrightarrow{x} \\ \end{matrix}\terminate{array} \)

Types of Simple Harmonic Motility

SHM or Elementary Harmonic Move can be classified into two types:

• Linear SHM
• Angular SHM

Linear Simple Harmonic Motility

When a particle moves to and fro almost a stock-still point (called equilibrium position) along with a straight line so its motion is called linear Uncomplicated Harmonic Motion.

For Example: spring-mass organization

Conditions for Linear SHM:

The restoring force or acceleration acting on the particle should always be proportional to the displacement of the particle and directed towards the equilibrium position.

\(\begin{array}{l}\brainstorm{matrix} \overrightarrow{F}\propto -\overrightarrow{x} \\ \overrightarrow{a}\,\,\propto -\overrightarrow{ten} \\ \terminate{matrix}\end{assortment} \)

• \(\begin{array}{50}\overrightarrow{x}\end{array} \)

  – deportation of particle from equilibrium position.

• \(\begin{assortment}{fifty}\overrightarrow{F}\terminate{array} \)

  – Restoring force

• \(\begin{array}{l}\overrightarrow{a}\,\terminate{array} \)

  - dispatch

Angular Uncomplicated Harmonic Motion

When a system oscillates angular long with respect to a fixed axis and then its move is called athwart elementary harmonic motion.

Conditions to Execute Athwart SHM:

The restoring torque (or) Athwart dispatch acting on the particle should always exist proportional to the angular deportation of the particle and directed towards the equilibrium position.

Τ ∝ θ or α ∝ θ

Where,

• Τ – Torque
• α angular dispatch
• θ – athwart displacement

Uncomplicated Harmonic Motion Fundamental Terms

Mean Position

The indicate at which net force acting on the particle is zero.

From the mean position, the strength interim on the particle is,

\(\begin{array}{l}\overrightarrow{F}\propto -\overrightarrow{x}\finish{array} \)

\(\begin{array}{l}\overrightarrow{a}\,\propto -\overrightarrow{x}\end{assortment} \)

Conditions at Hateful Position

\(\brainstorm{assortment}{l}\overrightarrow{{{F}_{net}}}=0\terminate{array} \)

.

\(\begin{array}{l}\overrightarrow{a}=0\end{assortment} \)

The force acting on the particle is negative of the displacement. So, this point of equilibrium volition exist a stable equilibrium.

Aamplitude in SHM

It is the maximum deportation of the particle from the hateful position.

Time Period and Frequency of SHM

The minimum time afterward which the particle keeps on repeating its motion is known as the time menstruation (or) the shortest time taken to consummate one oscillation is too defined as the time menstruum.

T = 2π/ω

Frequency:The number of oscillations per 2nd is defined as the frequency.

Frequency = 1/T and, athwart frequency ω = 2πf = 2π/T

Phase in SHM

The phase of a vibrating particle at whatsoever instant is the country of the vibrating (or) oscillating particle regarding its deportation and management of vibration at that detail instant.

The expression, position of a particle as a function of fourth dimension.

ten = A sin (ωt + Φ)

Where (ωt + Φ) is the phase of the particle, the phase angle at time t = 0 is known as the initial stage.

Phase Difference

The departure of total phase angles of 2 particles executing simple harmonic motility with respect to the mean position is known equally the phase difference. Two vibrating particles are said to be in the same phase, the phase difference between them is an even multiple of π.

ΔΦ = nπ where northward = 0, ane, ii, 3, . . . . .

Two vibrating particles are said to exist in opposite phase if the stage difference between them is an odd multiple of π.

ΔΦ = (2n + one) π where n = 0, 1, 2, 3, . . . . .

Simple Harmonic Motion Equation and its Solution

Consider a particle of mass (m) executing Simple Harmonic Motion along a path 10 o 10; the mean position at O. Let the speed of the particle exist v₀ when it is at position p (at a distance no from O)

At t = 0 the particle at P (moving towards the right)

At t = t the particle is at Q (at a distance x from O)

With a velocity (v)

The restoring forcefulness

\(\begin{array}{l}\overrightarrow{F}\finish{array} \)

at Q is given by

⇒

\(\brainstorm{array}{fifty}\overrightarrow{F}=-M\overrightarrow{x}\end{array} \)

Chiliad – is positive constant

⇒

\(\begin{array}{l}\overrightarrow{F}=m\overrightarrow{a}\terminate{assortment} \)

\(\begin{array}{l}\overrightarrow{a}\cease{array} \)

- dispatch at Q

⇒

\(\begin{array}{fifty}1000\overrightarrow{a}=-Yard\overrightarrow{10}\end{array} \)

⇒

\(\begin{array}{l}\overrightarrow{a}=-\left( \frac{K}{k} \right)\overrightarrow{x}\stop{array} \)

Put,

\(\brainstorm{array}{fifty}\frac{Chiliad}{thousand}={{\omega }^{2}}\end{array} \)

⇒

\(\begin{array}{l}\omega =\sqrt{\frac{K}{thou}}\stop{array} \)

⇒

\(\begin{array}{l}\overrightarrow{a}=-\left( \frac{Thousand}{m} \right)\overrightarrow{m}=-{{\omega }^{2}}\overrightarrow{x}\end{array} \)

Since,

\(\brainstorm{array}{50}\left[ \overrightarrow{a}=\frac{{{d}^{2}}ten}{d{{t}^{2}}} \correct]\stop{assortment} \)

\(\begin{array}{l}\frac{{{d}^{2}}\overrightarrow{x}}{d{{t}^{2}}}=-{{\omega }^{two}}\overrightarrow{x}\terminate{assortment} \)

d²x/dt² + ω²10 = 0, which is the differential equation for linear simple harmonic motility.

Solutions of Differential Equations of SHM

The differential equation for the Simple harmonic motion has the following solutions:

• \(\begin{array}{l}ten=A\sin \omega \,t\end{array} \)

  (This solution when the particle is in its mean position signal (O) in figure (a)

• \(\begin{assortment}{50}{{x}_{0}}=A\sin \phi\terminate{array} \)

  (When the particle is at the position & (not at mean position) in effigy (b)

• \(\begin{array}{50}x=A\sin \left( \omega t+\phi \right)\end{array} \)

  (When the particle at Q at in effigy (b) (whatever time t).

These solutions can exist verified by substituting this ten values in the above differential equation for the linear simple harmonic motility.

Angular Simple Harmonic Motion

A body free to rotate about an axis tin can make angular oscillations. For example, a photo frame or a agenda suspended from a nail on the wall. If it is slightly pushed from its hateful position and released, it makes angular oscillations.

Conditions for an Angular Oscillation to exist Angular SHM

The body must experience a cyberspace Torque that is restoring in nature. If the angle of oscillation is small, this restoring torque will exist directly proportional to the athwart displacement.

Τ ∝ – θ

Τ = – kθ

Τ = Iα

α = – kθ

\(\begin{array}{fifty}I\frac{{{d}^{ii}}\theta }{d{{t}^{2}}}=-K\theta\end{assortment} \)

\(\begin{array}{l}\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=-\left( \frac{K}{I} \right)\theta =-\omega _{0}^{2}\theta\terminate{array} \)

\(\begin{assortment}{fifty}\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=-\omega _{0}^{2}\theta =0\end{array} \)

This is the differential equation of an angular Uncomplicated Harmonic Motion. Solution of this equation is angular position of the particle with respect to time.

\(\begin{assortment}{l}\theta ={{\theta }_{0}}\sin \left( {{\omega }_{0}}t+\phi \right)\end{array} \)

So athwart velocity,

\(\begin{array}{50}\omega ={{\theta }_{0}}.\,{{\omega }_{0}}\cos \left( {{\omega }_{0}}t+\phi \right)\end{array} \)

θ₀ – amplitude of the angular SHM

Example:

• Simple pendulum
• Seconds pendulum
• The physical pendulum
• Torsional pendulum

Quantitative Analysis of SHM

Allow u.s. consider a particle executing Simple Harmonic Motion betwixt A and A₁ about passing through the mean position (or) equilibrium position (O). Its analysis is as follows

SHM about Position O

Displacement	x = -A	x = 0	x = +A
Acceleration	|a| = Max	a = 0	|a| = max
Speed	|v| = 0	|v| = Max	|v| = 0
Kinetic energy	KE = 0	KE = Max	KE = 0
Potential free energy	PE = Max	PE = Min	PE = Max

Equation of Position of a Particle equally a Role of Time

Allow us consider a particle, which is executing SHM at fourth dimension t = 0, the particle is at a distance from the equilibrium position.

Necessary weather for Simple Harmonic Motion

• \(\begin{array}{l}\overrightarrow{F}\propto -\overrightarrow{10}\terminate{array} \)

• \(\begin{array}{fifty}\overrightarrow{a}\propto -\overrightarrow{x}\end{array} \)

• \(\begin{assortment}{l}\overrightarrow{a}=-{{\omega }^{two}}ten\cease{array} \)

• \(\begin{array}{l}\overrightarrow{a}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}\end{array} \)

• \(\begin{array}{l}\overrightarrow{a}=v\frac{dv}{dx}=-{{\omega }^{ii}}x\stop{array} \)

• \(\begin{array}{l}\int\limits_{0}^{5}{vdv}=\int\limits_{0}^{x}{-{{\omega }^{ii}}xdx}\stop{array} \)

• \(\brainstorm{array}{fifty}\frac{{{v}^{2}}}{2}=\frac{-{{\omega }^{2}}{{x}^{2}}}{2}+c\stop{assortment} \)

  … (1)

Some condition we know:

At point A v = 0 [x = A] the equation (1) becomes

\(\brainstorm{array}{fifty}\frac{{{v}^{two}}}{2}=\frac{-{{\omega }^{ii}}{{A}^{2}}}{2}+c\cease{array} \)

v = 0

O =

\(\begin{array}{50}\frac{-{{\omega }^{2}}{{A}^{2}}}{2}+c\terminate{array} \)

c =

\(\begin{array}{l}\frac{{{\omega }^{two}}{{A}^{2}}}{2}\stop{array} \)

Sub the value of C in equation (i)

\(\brainstorm{array}{l}\frac{{{five}^{2}}}{2}=\frac{-{{\omega }^{2}}{{x}^{two}}}{2}+\frac{{{\omega }^{2}}{{A}^{ii}}}{ii}\end{assortment} \)

⇒

\(\begin{array}{l}{{v}^{two}}=-{{\omega }^{two}}{{x}^{2}}+{{\omega }^{2}}{{A}^{2}}\end{array} \)

⇒

\(\begin{array}{fifty}{{five}^{2}}={{\omega }^{2}}\left( {{A}^{ii}}-{{x}^{2}} \right)\end{array} \)

v =

\(\begin{array}{fifty}\sqrt{{{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right)}\end{array} \)

v =

\(\begin{array}{l}\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\finish{array} \)

… (2)

where, v is the velocity of the particle executing simple harmonic movement from definition instantaneous velocity

v =

\(\begin{array}{l}\frac{dx}{dt}=\omega \sqrt{{{A}^{2}}-{{10}^{two}}}\end{array} \)

⇒

\(\begin{assortment}{l}\int{\frac{dx}{\sqrt{{{A}^{two}}-{{10}^{2}}}}}=\int\limits_{0}^{t}{\omega dt}\end{array} \)

⇒

\(\brainstorm{array}{l}{{\sin }^{-ane}}\left( \frac{x}{A} \correct)=\omega t+\phi\end{array} \)

x = Asin ( ωt + Φ) . . . . . (3)

The equation (3) – equation of position of a particle as a function of time.

Case 1: If at t = 0

The particle at x = x0

⇒

\(\begin{array}{50}{{\sin }^{-ane}}\left( \frac{ten}{A} \correct)=\omega t+\phi\end{assortment} \)

i.e.

\(\begin{array}{l}{{\sin }^{-one}}\left( \frac{{{x}_{0}}}{A} \right)=\phi\finish{array} \)

initial stage of the particle

Φ is the initial stage of the particle

Case 2: If at t = 0

The particle at x = 0

\(\begin{array}{l}{{\sin }^{-i}}\left( \frac{O}{A} \right)=\phi\end{array} \)

i.e. Φ = 0

Case 3:If the particle is at 1 of its extreme position x = A at t = 0

⇒

\(\brainstorm{array}{l}{{\sin }^{-i}}\left( \frac{A}{A} \right)=\phi\end{array} \)

⇒

\(\begin{array}{l}{{\sin }^{-1}}\left( 1 \right)=\phi\terminate{array} \)

⇒ π/ii = Φ

Then, the value can exist anything depending upon the position of the particle at t = 0. That is why it is called initial phase of the particle.

Now if we run across the equation of position of the particle with respect to time

π/2 = 10 = A sin ( ωt + Φ)

sin (ωt + Φ) – is the periodic function, whose menstruation is T = 2π/ω

Which can be anything sine function or cosine part

Time Period of SHM

The coefficient of t is ω.

So the time menstruum T = 2π/ω

ω =2π/T = 2πf

ωt = angular frequency of SHM.

From the expression of particle position as a part of time:

We tin can detect particles, deportation

\(\begin{array}{l}\left( \overrightarrow{x} \correct), \cease{array} \)

velocity

\(\begin{array}{50}\left( \overrightarrow{v} \right)\end{array} \)

and dispatch as follows.

Velocity of a particle executing Simple Harmonic Motion

Velocity in SHM is given by five = dx/dt,

x = A sin (ωt + Φ)

five =

\(\begin{array}{l}\frac{d}{dt}A\sin \left( \omega t+\phi \right)=\omega A\cos \left( \omega t+\phi \right)\terminate{array} \)

v =

\(\begin{array}{l}A\omega \sqrt{i-{{\sin }^{2}}\omega t}\terminate{array} \)

Since, x = A sin ωt

\(\begin{array}{l}\frac{{{ten}^{2}}}{{{A}^{2}}}={{\sin }^{ii}}\omega \,t\end{array} \)

⇒

\(\begin{array}{l}v = A\omega \sqrt{1-\frac{{{x}^{2}}}{{{A}^{2}}}}\cease{array} \)

⇒

\(\begin{array}{l}v = \omega \sqrt{{{A}^{2}}-{{x}^{ii}}}\stop{array} \)

On squaring both sides

⇒

\(\begin{array}{50}{{v}^{2}}={{\omega }^{ii}}\left( {{A}^{2}}-{{x}^{2}} \right)\terminate{array} \)

⇒

\(\begin{array}{l}\frac{{{v}^{2}}}{{{\omega }^{two}}}=\left( {{A}^{two}}-{{10}^{two}} \right)\end{assortment} \)

⇒

\(\begin{array}{l}\frac{{{v}^{ii}}}{{{\omega }^{2}}{{A}^{2}}}=\left( ane-\frac{{{10}^{2}}}{{{A}^{ii}}} \right)\end{array} \)

⇒

\(\begin{array}{l}\frac{{{v}^{ii}}}{{{A}^{2}}}+\frac{{{v}^{2}}}{{{A}^{2}}{{\omega }^{2}}}=1\terminate{array} \)

this is an equation of an ellipse.

The curve between deportation and velocity of a particle executing the simple harmonic motion is an ellipse.

When ω = 1 then, the curve betwixt v and x will be circular.

Acceleration in SHM

\(\begin{array}{50}\overrightarrow{a}=\frac{dv}{dt}=\frac{d}{dt}\left( A\omega \cos \omega t+\phi \right)\end{array} \)

⇒

\(\begin{array}{l}\overrightarrow{a}=-{{\omega }^{2}}A\sin \left( \omega t+\phi \right)\end{array} \)

⇒

\(\begin{array}{50}\left| a \right|=-{{\omega }^{2}}ten\end{assortment} \)

Hence the expression for displacement, velocity and acceleration in linear simple harmonic motion are

• x = A sin ( ωt + Φ)
• v =

  \(\begin{array}{l}A\omega \cos \left( \omega t+\phi \right)=\omega \sqrt{{{A}^{2}}-{{x}^{ii}}}\end{array} \)

  and
• a =

  \(\begin{assortment}{l}-A{{\omega }^{2}}\sin \left( \omega t+\phi \right)=-{{\omega }^{ii}}x\end{array} \)

Energy in Simple Harmonic Motion (SHM)

The system that executes SHM is chosen the harmonic oscillator.

Consider a particle of mass m, executing linear simple harmonic motion of angular frequency (ω) and amplitude (A) the displacement

\(\brainstorm{assortment}{l}\left( \overrightarrow{x} \correct),\stop{array} \)

velocity

\(\begin{array}{l}\left( \overrightarrow{v} \right)\stop{array} \)

and dispatch

\(\brainstorm{assortment}{l}\left( \overrightarrow{a} \correct)\terminate{array} \)

at any time t are given by

x = A sin (ωt + Φ)

v =

\(\begin{assortment}{fifty}A\omega \cos \left( \omega t+\phi \right)=\omega \sqrt{{{A}^{ii}}-{{x}^{2}}}\end{array} \)

a =

\(\begin{array}{l}-{{\omega }^{2}}A\sin \left( \omega t+\phi \right)=-{{\omega }^{2}}x\end{array} \)

The restoring force

\(\begin{array}{l}\left( \overrightarrow{F} \right)\terminate{array} \)

acting on the particle is given by

F = -kx where one thousand = mω².

Kinetic Energy of a Particle in SHM

Kinetic Energy =

\(\begin{array}{l}\frac{1}{ii}m{{five}^{two}}\end{assortment} \)

\(\begin{array}{l}\left[ Since, \;{{v}^{ii}}={{A}^{2}}{{\omega }^{ii}}{{\cos }^{2}}\left( \omega t+\phi \right) \right]\end{array} \)

=

\(\begin{array}{50}\frac{1}{ii}m{{\omega }^{2}}{{A}^{2}}{{\cos }^{2}}\left( \omega t+\phi \right)\cease{array} \)

=

\(\begin{array}{l}\frac{ane}{2}m{{\omega }^{ii}}\left( {{A}^{2}}-{{x}^{2}} \right)\end{assortment} \)

Therefore, the Kinetic Energy =

\(\begin{array}{l}\frac{1}{ii}m{{\omega }^{two}}{{A}^{2}}{{\cos }^{two}}\left( \omega t+\phi \right)=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{two}}-{{x}^{2}} \right)\stop{assortment} \)

Potential Energy of SHM

The total work washed by the restoring forcefulness in displacing the particle from (x = 0) (hateful position) to x = x:

When the particle has been displaced from x to x + dx the work done by restoring strength is

dw = F dx = -kx dx

w =

\(\begin{array}{l}\int{dw}=\int\limits_{0}^{x}{-kxdx=\frac{-k{{x}^{2}}}{two}}\end{array} \)

=

\(\begin{array}{l}-\frac{m{{\omega }^{two}}{{x}^{2}}}{2}\stop{array} \)

\(\begin{array}{l}\left[ \,k=m{{\omega }^{ii}} \right]\end{array} \)

=

\(\begin{array}{fifty}-\frac{thousand{{\omega }^{2}}}{2}{{A}^{two}}{{\sin }^{2}}\left( \omega t+\phi \right)\end{array} \)

Potential Energy = -(piece of work done by restoring force)

Potential Energy =

\(\begin{array}{l}\frac{m{{\omega }^{two}}{{x}^{2}}}{2}=\frac{thou{{\omega }^{2}}{{A}^{ii}}}{two}{{\sin }^{2}}\left( \omega t+\phi \right)\stop{assortment} \)

Total Mechanical Energy of the Particle Executing SHM

E = KE + PE

E =

\(\begin{array}{50}\frac{i}{ii}m{{\omega }^{2}}\left( {{A}^{2}}-{{x}^{ii}} \right)+\frac{one}{2}1000{{\omega }^{two}}{{x}^{ii}}\end{array} \)

Eastward =

\(\begin{array}{fifty}\frac{1}{2}1000{{\omega }^{two}}{{A}^{2}}\end{assortment} \)

Hence the total energy of the particle in SHM is constant and it is independent of the instantaneous displacement.

⇒ Human relationship between Kinetic Free energy, Potential Energy and time in Simple Harmonic Motion at t = 0, when x = ±A.

⇒ Variation of Kinetic Energy and Potential Energy in Simple Harmonic Movement with deportation:

Geometrical Interpretation of Simple Harmonic Motion

If a particle is moving with compatible speed along the circumference of a circle then the straight line motion of the foot of the perpendicular drawn from the particle on the diameter of the circle is called elementary harmonic motility.

SHM every bit a Project of Circular Move

The particle is at position P at t = 0 and revolves with a abiding angular velocity (ω) along a circumvolve. The project of P on the bore along the 10-axis (G). At the later time (t) the particle is at Q. At present its project on the bore along the ten-axis is North.

As the particle P revolves around in a circle anti-clockwise its projection Thousand follows it up moving dorsum and along along the diameter such that the deportation of the signal of projection at any fourth dimension (t) is the x-component of the radius vector (A).

x = A cos ( ωt + Φ) . . . . . . . (1)

y = A sin ( ωt + Φ) . . . . . (two)

Thus, we run into that the uniform circular motion is the combination of two mutually perpendicular linear harmonic oscillation.

It implies that P is under uniform circular motion, (M and N) and (K and L) are performing simple harmonic motion almost O with the aforementioned angular speed ω every bit that of P.

P is under uniform circular movement, which volition have centripetal acceleration along A (radius vector)

\(\brainstorm{array}{l}\overrightarrow{{{a}_{c}}}=A{{\omega }^{2}}\end{array} \)

(towards the centre)

Information technology can be resolved into ii components:

• \(\begin{array}{fifty}{{a}_{N}}=A{{\omega }^{2}}{{\sin }^{2}}\left( \omega t+\phi \right)\terminate{array} \)

• \(\begin{array}{l}{{a}_{Fifty}}=A{{\omega }^{2}}{{\cos }^{2}}\left( \omega t+\phi \right)\stop{assortment} \)

a_N and a_L acceleration corresponding to the points Due north and L respectively.

In the above discussion, the pes of projection on the x-axis is called horizontal phasor.

Similarly, the foot of the perpendicular on the y-axis is called vertical phasor. Already nosotros know the vertical and horizontal phasor will execute the unproblematic harmonic motion of amplitude A and angular frequency ω. The phases of the two SHM differ by π/2.

Problem-Solving Strategy in Horizontal Phasor

1. Permit us presume a circle of radius equal to the amplitude of SHM.
2. Assume a particle rotating in a circular path moving with constant same equally that of simple harmonic motion in the clockwise direction.
3. Bending made by the particle at t = 0 with the upper vertical axis is equal to φ (stage abiding).
4. The horizontal component of the velocity of a particle gives you the velocity of a particle performing the simple harmonic movement.
5. The component of the acceleration of a particle in the horizontal direction is equal to the acceleration of the particle performing SHM. [In uniform circular acceleration centripetal but a_c = ω²A].

Waves and Simple Harmonic Motion Important Questions

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