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# derive an expression for displacement velocity and acceleration of a particle executing shm

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## Mention the expression for velocity and acceleration of a particle executing SHM.

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## Mention the expression for velocity and acceleration of a particle executing SHM.

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Updated on : 2022-09-05

The velocity of the particle executing SHM at any instant of time is given by

Solution Verified by Toppr v=Aωcosωt=ω A 2 −y 2 ​

The acceleration of the particle executing SHM is given by

a= dt dv ​ =−Aω 2 sinωt=−ω 2 y

where ω - angular frequency, A - amplitude, y - displacement of particle from its mean position in time t .

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## Derive the expressions for displacement velocity and acceleration of a particle executes S.H.M.

Displacement of a body in S.H.M. X=A cos(omegat+phi). i) Displacement(x): At t=0 displacement x=A i.e., extreme position when omegat=phi=90^(@) displacement x=0 at mean position at any point x=Acos(omegat+phi). ii) Velocity (V): Velocity of a body in S.H.M. is V=(dx)/(dt)=(d)/(dt){Aomegacos(omegat+phi)} therefore V=-A omegasin(omegat+phi) or V=-omegasqrt(A^(2)-x^(2)) When (omegat=phi)=0 then velocity v=0. For points where (omegat+phi)=90^(@) Velocity V=-Aomegai.e., velocity is maximum. iii)Acceleration (a): Acceleration of a body in S.H.M. is a=(dy)/(dx) =d/dt(-Aomegasin(omegat+phi)=-Aomega^(2)cos(omegat+phi)=-omega^(2)x) a(max)=-omega^(2)A.

Derive the expressions for displacement velocity and acceleration of a particle executes S.H.M.

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Updated On: 27-06-2022

Text Solution Solution

Displacement of a body in S.H.M.

X = A cos ( ω t + ϕ ) .

i) Displacement(x): At

t = 0 displacement x = A

i.e., extreme position when

ω t = ϕ = 90 ∘ displacement x = 0

at mean position at any point

x = A cos ( ω t + ϕ ) .

ii) Velocity (V): Velocity of a body in S.H.M. is

V = d x d t = d d t { A ω cos ( ω t + ϕ ) } ∴ V = − A ω sin ( ω t + ϕ ) or V = − ω √ A 2 − x 2 When ( ω t = ϕ ) = 0 then velocity v = 0 . For points where ( ω t + ϕ ) = 90 ∘ Velocity V = − A ω

i.e., velocity is maximum.

iii)Acceleration (a): Acceleration of a body in S.H.M. is

a = d y d x = d d t ( − A ω sin ( ω t + ϕ ) = − A ω 2 cos ( ω t + ϕ ) = − ω 2 x ) a max = − ω 2 A . Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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## Mention the expression for velocity and acceleration of a particle executing SHM.

Mention the expression for velocity and acceleration of a particle executing SHM.. Ans: Hint:To find the expressions of velocity and acceleration of a particle executing simple harmonic motion, we need to use the equation of displacement of the parti...

## Mention the expression for velocity and acceleration of a particle executing SHM.

Last updated date: 14th Mar 2023

• Total views: 205.5k • Views today: 2.86k Answer Verified 205.5k+ views

Hint:To find the expressions of velocity and acceleration of a particle executing simple harmonic motion, we need to use the equation of displacement of the particle from its equilibrium position in simple harmonic motion. The first derivative of the displacement with respect to time will give us the expression of velocity and by further differentiation, the second derivative of the displacement with respect to time will give us the expression of acceleration.Formulas used:

x=Asin(ωt+ϕ) x=Asin⁡(ωt+ϕ) , Where, x x

is the displacement from equilibrium position,

A A is amplitude, ω ω

is angular frequency,

t t is time and ϕ ϕ

is initial face angle. Here,

(ωt+ϕ) (ωt+ϕ) is called phase. v= dx dt v=dxdt , Where, v v

is the velocity of particle,

x x

is the displacement from equilibrium position and

t t is time a= dv dt a=dvdt , Where, a a

is the acceleration of the particle,

v v

is the velocity of particle and

t t is time

We will first derive the expression for the velocity of the particle which executes simple harmonic motion by differentiating its displacement from the equilibrium position with respect to time. We know that the displacement from equilibrium position is given by:

x=Asin(ωt+ϕ) x=Asin⁡(ωt+ϕ) v= dx dt v=dxdt ⇒v= d dt

(Asin(ωt+ϕ))=ωAcos(ωt+ϕ)

⇒v=ddt(Asin⁡(ωt+ϕ))=ωAcos⁡(ωt+ϕ)

Now, at the equilibrium, initial phase angle is zero which means

ϕ=0 ϕ=0 ⇒v=ωAcosωt=ωA 1− sin 2 ωt − − − − − − − − √

⇒v=ωAcos⁡ωt=ωA1−sin2ωt

Also at equilibrium,

x=Asinωt⇒ sin 2 ωt= x 2 A 2

x=Asin⁡ωt⇒sin2ωt=x2A2

Putting this value in the equation of velocity, we get

v=Aω 1− x 2 A 2 − − − − − − √ =ω A 2 − x 2 − − − − − − √ v=Aω1−x2A2=ωA2−x2

Thus, the expression of the velocity of the particle executes simple harmonic motion is

v=ω A 2 − x 2 − − − − − − √ v=ωA2−x2

Now, let us find the expression of acceleration of the particle executing simple harmonic motion by differentiating its velocity with respect to time.

a= dv dt a=dvdt ⇒a= d dt (ωAcos(ωt+ϕ))=− ω 2 Asin(ωt+ϕ)

⇒a=ddt(ωAcos⁡(ωt+ϕ))=−ω2Asin⁡(ωt+ϕ)

Thus, the expressions for velocity and acceleration of a particle executing simple harmonic motion are

But, we know that x=Asin(ωt+ϕ) x=Asin⁡(ωt+ϕ) ∴a=− ω 2 x ∴a=−ω2x v=ω A 2 − x 2 − − − − − − √ v=ωA2−x2 and a=− ω 2 x a=−ω2x respectively.

v=ω A 2 − x 2 − − − − − − √ v=ωA2−x2

Squaring both sides, we get

v 2 = ω 2 ( A 2 − x 2 ) ⇒ v 2 ω 2 = A 2 − x 2 ⇒ v 2 ω 2 A 2 =1− x 2 A 2 ⇒ x 2 A 2 + v 2 ω 2 A 2 =1

v2=ω2(A2−x2)⇒v2ω2=A2−x2⇒v2ω2A2=1−x2A2⇒x2A2+v2ω2A2=1

This is the equation of an ellipse.

Therefore, we can say that the curve between displacement and velocity of a particle executing the simple harmonic motion is an ellipse.

Also, the acceleration of the particle in simple harmonic motion is given by

a=− ω 2 x a=−ω2x

. And if we plot the graph of acceleration and displacement, it will be a straight line.

Note:Here, we have seen that the displacement of the particle executing SHM is dependent on amplitude, angular frequency, time and initial face angle. We have derived the formula for velocity and acceleration of this particle using its displacement. That is why they both are also dependent on the same parameters.

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