Find expression for velocity and acceleration in SHM . What is the phase relationship between them?

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atul nagar · Answer 1 · 2019-06-11T08:55:39+0000

For SHM

Displacement X= A Sin wt

Velocity V= Aw Cos wt

Acceleration a = - Aw²Sin wt

a = -Aw² √(1-V²/w²A²)

a= -w√(w²A² - V²)

anukriti · Answer 2 · 2019-06-11T09:14:21+0000

We know that velocity of a particle is given by
v=dx/dt
In SHM displacement of particle is given by
x=A cos(ωt+φ)
now differentiating it with respect to t
v=dx/dt= Aω(-sin(ωt+φ)) (1)
Here in equation 1 quantity Aω is known as velocity amplitude and velocity of oscillating particle varies between the limits ±ω.
From trignometry we know that
     cos²θ + sin²θ=1
⇒
     A² sin²(ωt+φ)= A²- A²cos²(ωt+φ)
Or
     sin(ωt+φ)=[1-x²/A²]            (2)
putting this in equation 1 we get,

---- (3)
From this equation 3 we notice that when the displacement is maximum i.e. ±A the velocity v=0, because now the oscillator has to return to change its direction.
Figure below shows the variation of velocity with time in SHM with initial phase φ=0.

Again we know that acceleration of a particle is given by
a=dv/dt
where v is the velocity of particle executing motion.
In SHM velocity of particle is give by,
v= -ωsin(ωt+φ)
differentiating this we get,

or,
a=-ω²Acos(ωt+φ) (4)
Equation 4 gives an acceleration of particle executing simple harmonic motion and quantity ω² is called acceleration amplitude and the acceleration of oscillating particle varies between the limits ±ω²A.
Putting equation x=A cos(ωt+φ) in 4 we get
a=-ω²x (5)
which shows that acceleration is proportional to the displacement but in opposite direction.
Thus from above equation we can see that when x is maximum (+A or -A), the acceleration is also maximum(-ω²A or +ω²A) but is directed in direction opposite to that of displacement.
Figure below shows the variation of acceleration of particle in SHM with time having initial phase φ=0.