Physics

$x=Asinwt$

By definition, $v=dtdx $

or, $v=Awcoswt$

Acceleration $a=dtdv $

or $a=−Aw_{2}sinwt$

or $a=−w_{2}x$

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

$a=−ω_{2}x$

or

$a=m−K x$

so, from graph

$−mK =−1$ $(∵slopeis−1)$

$mK =1$

Time period $=2πKm $

$=2π11 $

$=2π$

$x=Asinwt$

By definition, $v=dtdx $

or, $v=Awcoswt$ ... (1)

Since $sin_{2}wt+cos_{2}wt=1$

From equation (1).

$v=wA_{2}−x_{2} $

$x=Asinwt$

By definition, $v=dtdx $

or, $v=Awcoswt$

Acceleration is given by

$a=dtdv $

or $a=−Aw_{2}sinwt$

Solution:

From the image,

$ϕ=−θ+ωt$.............................................$(1)$

$180−ϕ=θ+ωt$......................................$(2)$

substituting value of $ϕ$ from equn $(1)$ in $(2)$, we will get $ωt=90$.

That means that particles will have to rotate by $90_{0}$ in order to be again at same displacement.

Hence from $(1),$

we get $ϕ=−θ+90$

now from figure,

$y=Asinθ$.....................................................$(3)$ and

$y_{′}=Asinϕ=Asin(−θ+90)=Acosθ=A1−sin_{2}θ $

From eqn $(3)$

$y_{′}=A_{2}−y_{2} $....................................................$(4)$

Initially magnitude of velocity, $=1.2m/s$

Therefore $1.2=ωA_{2}−y_{2} $......................$(5)$

and Finally $1.6=ωA_{2}−y_{′2} $

that gives, $1.6=ωy$ $($using eqn $(4))$ ......................$(6)$

Putting $(6)$ in eqn $(5)$ after simplifying,we will get ,

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

at mean position.

$a=−Aw_{2}sinwt$

From this expression one could infer directly that acceleration is maximum for:

$wt=90$ ( at the extreme position)

Acceleration is minimum for:

$wt=0$ ( at the mean position)

$x=Asinwt$

By definition, $v=dtdx $

or, $v=Awcoswt$ ... (1)

Clearly from equation (1) maximum velocity will be:

$v=Aw$ for $wt=0$

and

$v=0$ for $wt=90$

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

Angular acceleration as a function of time in SHM:

$α=−w_{2}θ_{o}sinwt$

Since,

$θ=θ_{o}sinwt$

$τ=−kθ$

General equation for angular displacement:

$θ=θ_{o}sin(wt+ϕ)$

Angular velocity $=dtdθ $

or, Angular velocity = $θ_{o}wcos(wt+ϕ)$

where $w=T2π $

Where $T$ is time period of SHM.

$a$ Vs $x$ plot essentially represents a straight line with slope $−w_{2}$ .

Direction of acceleration is opposite to the direction of displacement.

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