Question

The position x of a particle moving along x axis varies with time t as x = Asin(wt) , where A and w are constants . The acceleration of the particle varies with its position as ​

Answer 1

Given :-

• The position x of a particle moving along x axis varies with time t as $x = Asin(\omega t)$

To find :-

• The acceleration of the particle .

Solution :-

Here we are given that the position of the particle varies with time (t) as ,

$\implies x = A\ sin (\omega t)$

Differenciate both sides with respect to t , first order of differenciation will give velocity .

$\implies \dfrac{dx}{dt}=\dfrac{d}{dt}( A \ sin (\omega t) )$

As we know that $\dfrac{d}{dx}sin(ax) = a cos(ax)$ , where a is a constant . Hence,

$\implies v = A \dfrac{ d( sin (\omega t )) }{dt} \\\\\implies v = A . \omega cos (\omega t )$

Again differenciate both sides wrt t . Second order of differenciation will give acceleration .

$\implies \dfrac{dv}{dt}=\dfrac{d}{dt}( A \omega cos(\omega t) ) \\\\\implies a = A\omega\bigg[\dfrac{d}{dt}( \omega cos(\omega t) )\bigg] \\\\\implies a = A \omega.\omega . (-sin (\omega t))\\\\\implies a = \omega^2 . (- A sin(\omega t) )$

Now , since $x = A sin(\omega t)$ ,

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

Answer 2

Explanation:

Given :-

The position x of a particle moving along x axis varies with time t as x = Asin(\omega t)x=Asin(ωt)

To find :-

The acceleration of the particle .

Solution :-

Here we are given that the position of the particle varies with time (t) as ,

\implies x = A\ sin (\omega t)⟹x=A sin(ωt)

Differenciate both sides with respect to t , first order of differenciation will give velocity .

\implies \dfrac{dx}{dt}=\dfrac{d}{dt}( A \ sin (\omega t) )⟹

dt

dx

=

dt

d

(A sin(ωt))

As we know that \dfrac{d}{dx}sin(ax) = a cos(ax)

dx

d

sin(ax)=acos(ax) , where a is a constant . Hence,

\begin{gathered}\implies v = A \dfrac{ d( sin (\omega t )) }{dt} \\\\\implies v = A . \omega cos (\omega t ) \end{gathered}

⟹v=A

dt

d(sin(ωt))

⟹v=A.ωcos(ωt)

Again differenciate both sides wrt t . Second order of differenciation will give acceleration .

\begin{gathered}\implies \dfrac{dv}{dt}=\dfrac{d}{dt}( A \omega cos(\omega t) ) \\\\\implies a = A\omega\bigg[\dfrac{d}{dt}( \omega cos(\omega t) )\bigg] \\\\\implies a = A \omega.\omega . (-sin (\omega t))\\\\\implies a = \omega^2 . (- A sin(\omega t) )\end{gathered}

⟹

dt

dv

=

dt

d

(Aωcos(ωt))

⟹a=Aω[

dt

d

(ωcos(ωt))]

⟹a=Aω.ω.(−sin(ωt))

⟹a=ω

2

.(−Asin(ωt))

Now , since x = A sin(\omega t)x=Asin(ωt) ,

\implies\underline{\underline{ a = -\omega^2 x}}⟹

a=−ω

2

x