Physics, asked by Anonymous, 5 months ago

Using differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.​


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Answers

Answered by TheValkyrie
93

Answer:

\sf a =- \omega^{2}\:x

\sf v=\pm \omega\sqrt{A^2-x^2}

Explanation:

Given:

  • Differential Equation of linear S.H.M

To Find:

  • To obtain an expression for velocity in S.H.M
  • To obtain an expression for acceleration in S.H.M

Solution:

The differential equation for S.H.M is given by,

\sf \dfrac{d^2x}{dt^2} + \omega^{2}\:x=0

where ω² = k/m , k is the force constant and m is the mass of the particle.

We know that acceleration is the change in velocity of the body, and velocity is the change in displacement that is,

\sf v=\dfrac{dx}{dt}

\sf a=\dfrac{dv}{dt}

\sf a= \dfrac{d}{dt} (\dfrac{dx}{dt})

\sf a=\dfrac{d^2x}{dt^{2} }

Therefore,

\sf a + \omega^{2}\:x=0

\boxed{\sf a =- \omega^{2}\:x}

where x = displacement of the particle

We know,

\sf a= \dfrac{dv}{dt}

This can be written as,

\sf a=\dfrac{dv}{dx}\times \dfrac{dx}{dt}

\sf a=v\times \dfrac{dv}{dx}

But we know that the acceleration of a particle in SHM is,

a = -ω² x

Therefore,

\sf v\times \dfrac{dv}{dx} =-\omega^{2}\:x

\sf v\:dv=-\omega^{2}\:dx

By integration on both sides we get,

\displaystyle \sf \int\limits {v} \, dv=\int\limits {-\omega^2\:x} \, dx

Now we know that,

\displaystyle \sf \int\limits {x^n} \, dx =\dfrac{x^{n+1}}{n+1}

Therefore,

\sf \dfrac{v^2}{2} =-\omega^{2}\times \dfrac{x^2}{2}+C

When the particle reaches the extreme position, ie, when x = A, the velocity becomes 0.

Hence,

\sf 0=-\omega^2\times \dfrac{A^2}{2}+C

\sf C=\omega^2\times \dfrac{A^2}{2}

Substitute the value of C,

\sf \dfrac{v^2}{2} =-\omega^{2}\times \dfrac{x^2}{2}+\omega^{2}\times \dfrac{A^2}{2}

v² = -ω²x² + ω²A²

v² = ω² (A² - x²)

\boxed{\sf v=\pm \omega\sqrt{A^2-x^2}}


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Answered by DARLO20
33

\Large{\purple{\underline{\textsf{\textbf{Step-by-step\:Explanation\::}}}}} \\

Dɪғғʀɴɪʟ ғʀ ғ Lɪɴʀ S.H.M :

\red\checkmark\:\:{\underline{\blue{\boxed{\bf{\green{\dfrac{d^2x}{dt^2}\:+\:\omega^{2}\:x\:=\:0}}}}}} \\

Wʜʀ,

  • \bf{\omega\:=\:\sqrt{\dfrac{k}{m}}} \\ , is the angular frequency of the particle performing the linear simple harmonic motion.

W ɴ ʜ,

In any S.H.M, there will always be acting as a restoring force which will try to bring the object back to the mean position. This force will result in an acceleration in the object.

Tʜʀғʀ,

\longmapsto\:\:\bf\blue{F\:=\:-Kx}--(i) \\

[NOTE -ve sign indicates that the force will be opposite to the displacement.]

Wʜʀ,

  • F be the restoring force.

  • x be the displacement of the object from the mean position.

  • K is the force per unit displacement.

According to Newton's second law of motion,

\longmapsto\:\:\bf\pink{Force\:(F)\:=\:mass\:(m)\times{acceleration\:(a)}} \\

\longmapsto\:\:\bf{a\:=\:\dfrac{F}{m}} \\

➛ Putting the value of F from the equation (i), we get

\longmapsto\:\:\bf{a\:=\:\dfrac{-K}{m}\:x} \\

➛ From the differential form of the linear S.H.M,

  • ω = \bf{\sqrt{\dfrac{K}{m}}}

  • ω² = \bf{\dfrac{K}{m}}

➛ Putting the above value in the equation, we get

\longmapsto\:\:{\underline{\green{\boxed{\bf{\purple{a\:=\:-\omega^{2}\:x}}}}}}--(a) \\

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

\longrightarrow\:\:\bf{a\:=\:\dfrac{dv}{dt}} \\

Wʜʀ,

  • \bf{\dfrac{dv}{dt}} is the change in velocity w.r.t time.

➛ Thus, from equation (a)

:\implies\:\:\bf{\dfrac{dv}{dt}\:=\:-\omega^{2}\:x} \\

:\implies\:\:\bf{\dfrac{dv}{dx}\times{\dfrac{dx}{dt}}\:=\:-\omega^{2}\:x} \\

Wʜᴇʀᴇ,

  • \bf{\dfrac{dx}{dt}} is the change in displacement w.r.t time, i.e. velocity (v).

:\implies\:\:\bf{v\dfrac{dv}{dx}\:=\:-\omega^{2}\:x} \\

:\implies\:\:\bf{v\:{dv}\:=\:(-\omega^{2}\:x)\:dx} \\

↝ Let us integrates this equation as,

:\implies\:\:\bf{\int{v\:{dv}}\:=\:\int{(-\omega^{2}\:x)}\:dx} \\

:\implies\:\:\bf{\dfrac{v^2}{2}\:=\:\dfrac{-\omega^{2}\:x^2}{2}\:+\:C} \\

↝ In the above equation, when displacement is maximum,

  • x = A (amplitude)

↝ And at extreme point,

  • v = 0

\bf{0\:=\:\dfrac{-\omega^{2}\:A^2}{2}\:+\:C} \\

C = \rm{\dfrac{-\omega^{2}\:A^2}{2}\:} \\

↝ Let us substitute this in the equation as,

:\implies\:\:\bf{\dfrac{v^2}{2}\:=\:\dfrac{-\omega^{2}\:x^2}{2}\:+\:\dfrac{-\omega^{2}\:A^2}{2}} \\

:\implies\:\:\bf{v^2\:=\:\omega^{2}\:(A^2\:-\:x^2)} \\

:\implies\:\:{\underline{\pink{\boxed{\bf{\green{v\:=\:\pm\sqrt{\omega^{2}\:(A^2\:-\:x^2)}}}}}}} \\


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