Question

The ratio of maximum acceleration to maximum velocity of a particle performing simple harmonic motion is equal to--
A) amplitude
B) angular velocity
C) square of amplitude
D)square of angular velocity

Answer 1

The displacement of a particle executing simple harmonic motion (SHM) is given by,

$\longrightarrow\sf{x=A\sin\theta}$

where $\sf{A}$ is the amplitude and $\theta$ is the angular displacement.

But,

• $\sf{\theta=\omega t}$

where $\omega$ is the angular speed and $\sf{t}$ is the time. Hence the displacement is,

$\longrightarrow\sf{x=A\sin(\omega t)\quad\quad\dots(1)}$

Then the velocity will be,

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

$\longrightarrow\sf{v=\dfrac{d}{dt}[A\sin(\omega t)]}$

$\longrightarrow\sf{v=A\omega\cos(\omega t)}$

But since $\sf{\sin^2\alpha+\cos^2\alpha=1,}$

$\longrightarrow\sf{v=A\omega\sqrt{1-\sin^2(\omega t)}}$

$\longrightarrow\sf{v=\omega\sqrt{A^2-A^2\sin^2(\omega t)}}$

$\longrightarrow\sf{v=\omega\sqrt{A^2-\left(A\sin(\omega t)\right)^2}}$

From (1),

$\longrightarrow\sf{v=\omega\sqrt{A^2-x^2}\quad\quad\dots(2)}$

And the acceleration will be,

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

$\longrightarrow\sf{a=\dfrac{d}{dt}[A\omega\cos(\omega t)]}$

$\longrightarrow\sf{a=-A\omega^2\sin(\omega t)}$

From (1),

$\longrightarrow\sf{a=-\omega^2x}$

Negative sign shows that retardation occurs with displacement. However the magnitude of acceleration is,

$\longrightarrow\sf{|a|=\omega^2x\quad\quad\dots(3)}$

Maximum velocity is attained by the particle at mean position, i.e., when $\sf{x=0.}$

Then from (2),

$\longrightarrow\sf{v_{max}=\omega\sqrt{A^2-0^2}}$

$\longrightarrow\sf{v_{max}=A\omega}$

And maximum acceleration is attained by the particle at extreme positions, i.e., when $\sf{x=A.}$

Then from (3),

$\longrightarrow\sf{a_{max}=A\omega^2}$

Hence the ratio of maximum acceleration to maximum velocity is,

$\longrightarrow\sf{\dfrac{a_{max}}{v_{max}}=\dfrac{A\omega^2}{A\omega}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{a_{max}}{v_{max}}=\omega}}}$

I.e., the ratio is equal to angular velocity.

Answer 2

The displacement of a particle executing simple harmonic motion (SHM) is given by,

\longrightarrow\sf{x=A\sin\theta}⟶x=Asinθ

where \sf{A}A is the amplitude and \thetaθ is the angular displacement.

But,

\sf{\theta=\omega t}θ=ωt

where \omegaω is the angular speed and \sf{t}t is the time. Hence the displacement is,

\longrightarrow\sf{x=A\sin(\omega t)\quad\quad\dots(1)}⟶x=Asin(ωt)…(1)

Then the velocity will be,

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

dt

dx

\longrightarrow\sf{v=\dfrac{d}{dt}[A\sin(\omega t)]}⟶v=

dt

d

[Asin(ωt)]

\longrightarrow\sf{v=A\omega\cos(\omega t)}⟶v=Aωcos(ωt)

But since \sf{\sin^2\alpha+\cos^2\alpha=1,}sin

2

α+cos

2

α=1,

\longrightarrow\sf{v=A\omega\sqrt{1-\sin^2(\omega t)}}⟶v=Aω

1−sin

2

(ωt)

\longrightarrow\sf{v=\omega\sqrt{A^2-A^2\sin^2(\omega t)}}⟶v=ω

A

2

−A

2

sin

2

(ωt)

\longrightarrow\sf{v=\omega\sqrt{A^2-(A\sin(\omega t))^2}}⟶v=ω

A

2

−(Asin(ωt))

2

From (1),

\longrightarrow\sf{v=\omega\sqrt{A^2-x^2}\quad\quad\dots(2)}⟶v=ω

A

2

−x

2

…(2)

And the acceleration will be,

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

dt

dv

\longrightarrow\sf{a=\dfrac{d}{dt}[A\omega\cos(\omega t)]}⟶a=

dt

d

[Aωcos(ωt)]

\longrightarrow\sf{a=-A\omega^2\sin(\omega t)}⟶a=−Aω

2

sin(ωt)

From (1),

\longrightarrow\sf{a=-\omega^2x}⟶a=−ω

2

x

Negative sign shows that retardation occurs with displacement. However the magnitude of acceleration is,

\longrightarrow\sf{|a|=\omega^2x\quad\quad\dots(3)}⟶∣a∣=ω

2

x…(3)

Maximum velocity is attained by the particle at mean position, i.e., when \sf{x=0.}x=0.

Then from (2),

\longrightarrow\sf{v_{max}=\omega\sqrt{A^2-0^2}}⟶v

max

=ω

A

2

−0

2

\longrightarrow\sf{v_{max}=A\omega}⟶v

max

=Aω

And maximum acceleration is attained by the particle at extreme positions, i.e., when \sf{x=A.}x=A.

Then from (3),

\longrightarrow\sf{a_{max}=A\omega^2}⟶a

max

=Aω

2

Hence the ratio of maximum acceleration to maximum velocity is,

\longrightarrow\sf{\dfrac{a_{max}}{v_{max}}=\dfrac{A\omega^2}{A\omega}}⟶

v

max

a

max

=

Aω

Aω

2

\longrightarrow\sf{\underline{\underline{\dfrac{a_{max}}{v_{max}}=\omega}}}⟶

v

max

a

max

=ω

I.e., the ratio is equal to angular velocity.