# 1.2 Shm equation  (Page 2/2)

 Page 2 / 2

Angular acceleration, $\alpha =-\frac{mgl}{I}\theta$ ------ for rotational SHM of pendulum

where “m” and “I” are mass and moment of inertia of the oscillating objects in two systems.

In order to understand the nature of cause, we focus on the block-spring system. When the block is to the right of origin, “x” is positive and restoring spring force is negative. This means that restoring force (resulting from elongation of the spring) is directed left towards the neutral position (center of oscillation). This force accelerates the block towards the center. As a result, the block picks up velocity till it reaches the center.

The plot, here, depicts nature of force about the center of oscillation bounded between maximum displacements on either side.

As the block moves past the center, “x” is negative and force is positive. This means that restoring force (resulting from compression of the spring) is directed right towards the center. The acceleration is positive, but opposite to direction of velocity. As such restoring force decelerates the block.

In the nutshell, after the block is released at one extreme, it moves, first, with acceleration up to the center and then moves beyond center towards left with deceleration till velocity becomes zero at the opposite extreme. It is clear that block has maximum velocity at the center and least at the extreme positions (zero).

From the discussion, the characterizing aspects of the restoring force responsible for SHM are :

• The restoring force is always directed towards the center of oscillation.
• The restoring force changes direction across the center.
• The restoring force first accelerates the object till it reaches the center and then decelerates the object till it reaches the other extreme.
• The process of “acceleration” and “deceleration” keeps alternating in each half of the motion.

## Differential form of shm equation

We observed that acceleration of the object in SHM is proportional to negative of displacement. Here, we shall formulate the general equation for SHM in linear motion with the understanding that same can be extended to SHM along curved path. In that case, we only need to replace linear quantities with corresponding angular quantities.

$a=-{\omega }^{2}x$

where “ ${\omega }^{2}$ ” is a constant. The constant “ω” turns out to be angular frequency of SHM. This equation is the basic equation for SHM. For block-spring system, it can be seen that :

$\omega =\sqrt{\left(\frac{k}{m}\right)}$

where “k” is the spring constant and “m” is the mass of the oscillating block. We can, now, write acceleration as differential,

$\frac{{d}^{2}x}{d{t}^{2}}=-{\omega }^{2}x$

$⇒\frac{{d}^{2}x}{d{t}^{2}}+{\omega }^{2}x=0$

This is the SHM equation in differential form for linear oscillation. A corresponding equation of motion in the context of angular SHM is :

$⇒\frac{{d}^{2}\theta }{d{t}^{2}}+{\omega }^{2}\theta =0$

where "θ" is the angular displacement.

## Solution of shm differential equation

In order to solve the differential equation, we consider position of the oscillating object at an initial displacement ${x}_{0}$ at t =0. We need to emphasize that “ ${x}_{0}$ ” is initial position – not the extreme position, which is equal to amplitude “A”. Let

$t=0,\phantom{\rule{1em}{0ex}}x={x}_{0,}\phantom{\rule{1em}{0ex}}v={v}_{0}$

We shall solve this equation in two parts. We shall first solve equation of motion for the velocity as acceleration can be written as differentiation of velocity w.r.t to time. Once, we have the expression for velocity, we can solve velocity equation to obtain a relation for displacement as its derivative w.r.t time is equal to velocity.

## Velocity

We write SHM equation as differential of velocity :

$a=\frac{dv}{dt}==-{\omega }^{2}x$

$⇒\frac{dv}{dx}X\frac{dx}{dt}=-{\omega }^{2}x$

$⇒v\frac{dv}{dx}=-{\omega }^{2}x$

Arranging terms with same variable on either side, we have :

$⇒vdv=-{\omega }^{2}xdx$

Integrating on either side between interval, while keeping constant out of the integral sign :

$⇒\underset{{v}_{0}}{\overset{v}{\int }}vdv=-{\omega }^{2}\underset{{x}_{0}}{\overset{x}{\int }}xdx$

$⇒\left[\frac{{v}^{2}}{2}\underset{{v}_{0}}{\overset{v}{\right]}}=-{\omega }^{2}\left[\frac{{x}^{2}}{2}\underset{{x}_{0}}{\overset{x}{\right]}}$

$⇒\left({v}^{2}-{v}_{0}^{2}\right)=-{\omega }^{2}\left({x}^{2}-{x}_{0}^{2}\right)$

$⇒v=\sqrt{\left\{\left({v}_{0}^{2}+{\omega }^{2}{x}_{0}^{2}\right)-{\omega }^{2}{x}^{2}\right\}}$

$⇒v=\omega \sqrt{\left\{\left(\frac{{v}_{0}^{2}}{{\omega }^{2}}+{x}_{0}^{2}\right)-{x}^{2}\right\}}$

We put $\left(\frac{{v}_{0}^{2}}{{\omega }^{2}}+{x}_{0}^{2}\right)={A}^{2}$ . We shall see that “A” turns out to be the amplitude of SHM.

$⇒v=\omega \sqrt{\left({A}^{2}-{x}^{2}\right)}$

This is the equation of velocity. When x = A or -A,

$⇒v=\omega \sqrt{\left({A}^{2}-{A}^{2}\right)}=0$

when x = 0,

$⇒v\mathrm{max}=\omega \sqrt{\left({A}^{2}-{0}^{2}\right)}=\omega A$

## Displacement

We write velocity as differential of displacement :

$v=\frac{dx}{dt}=\omega \sqrt{\left({A}^{2}-{x}^{2}\right)}$

Arranging terms with same variable on either side, we have :

$⇒\frac{dx}{\sqrt{\left({A}^{2}-{x}^{2}\right)}}=\omega dt$

Integrating on either side between interval, while keeping constant out of the integral sign :

$⇒\underset{{x}_{0}}{\overset{x}{\int }}\frac{dx}{\sqrt{\left({A}^{2}-{x}^{2}\right)}}=\omega \underset{0}{\overset{t}{\int }}dt$

$⇒\left[{\mathrm{sin}}^{-1}\frac{x}{A}\underset{{x}_{0}}{\overset{x}{\right]}}=\omega t$

$⇒{\mathrm{sin}}^{-1}\frac{x}{A}-{\mathrm{sin}}^{-1}\frac{{x}_{0}}{A}=\omega t$

Let ${\mathrm{sin}}^{-1}\frac{{x}_{0}}{A}=\phi$ . We shall see that “φ” turns out to be the phase constant of SHM.

$⇒{\mathrm{sin}}^{-1}\frac{x}{A}=\omega t+\phi$

$⇒x=A\mathrm{sin}\left(\omega t+\phi \right)$

This is one of solutions of the differential equation. We can check this by differentiating this equation twice with respect to time to yield equation of motion :

$\frac{dx}{dt}=A\omega \mathrm{cos}\left(\omega t+\phi \right)$

$\frac{{d}^{2}x}{d{t}^{2}}=-A{\omega }^{2}\mathrm{sin}\left(\omega t+\phi \right)=-{\omega }^{2}x$

$⇒\frac{{d}^{2}x}{d{t}^{2}}+{\omega }^{2}x=0$

Similarly, it is found that equation of displacement in cosine form, $x=A\mathrm{cos}\left(\omega t+\phi \right)$ , also satisfies the equation of motion. As such, we can use either of two forms to represent displacement in SHM. Further, we can write general solution of the equation as :

$x=A\mathrm{sin}\omega t+B\mathrm{cos}\omega t$

This equation can be reduced to single sine or cosine function with appropriate substitution.

## Example

Problem 1: Find the time taken by a particle executing SHM in going from mean position to half the amplitude. The time period of oscillation is 2 s.

Solution : Employing expression for displacement, we have :

$x=A\mathrm{sin}\omega t$

We have deliberately used sine function to represent displacement as we are required to determine time for displacement from mean position to a certain point. We could ofcourse stick with cosine function, but then we would need to add a phase constant “π/2” or “-π/2”. The two approach yields the same expression of displacement as above.

Now, according to question,

$⇒\frac{A}{2}=A\mathrm{sin}\omega t$

$⇒\mathrm{sin}\omega t=\frac{1}{2}=\mathrm{sin}\frac{\pi }{6}$

$⇒\omega t=\frac{\pi }{6}$

$⇒t=\frac{\pi }{6\omega }=\frac{\pi T}{6X2\pi }=\frac{T}{12}=\frac{2}{12}=\frac{1}{6}\phantom{\rule{1em}{0ex}}s$

Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!