In the previous section, we saw simple harmonic motion (S.H.M). In this section, we will see velocity and acceleration in S.H.M.
Velocity in simple harmonic motion
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In fig.14.25 below, the magenta sphere is in uniform circular motion.
♦ Radius of the circle is A
♦ Angular speed of the sphere is $\small{\omega}$
 |
| Fig.14.25 |
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We know that, the "projection of the sphere on the x-axis" will be in
SHM between (−A,0) and (A,0). We want an expression which can be used
to calculate the "velocity of this projection" at any instant t.
This can be done in 4 steps:
1. Let at any instant t, the position of the sphere be P.
We
know that, at any instant, the tangential velocity of the sphere will
be $\small{\omega\,A}$. This is indicated by the white tangential arrow
at P, in the fig.14.25 above.
2. Now we drop two perpendicular green dashed lines, onto the x-axis.
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One through P and the other through the head of $\small{\omega\,A}$
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These two perpendicular lines will give the projection of $\small{\omega\,A}$ on the x-axis.
3. This projection is the velocity of the SHM at the instant t. It is denoted as $\small{v(t)}$. So our next aim is to find the expression for $\small{v(t)}$
4. Fig.14.26 below shows an enlarged view of the first quadrant.
 |
| Fig.14.26 |
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The white dotted line is parallel to the x-axis and it passes through P. So by the rule of alternate angles, angle between OP and the dotted line is $\small{(\omega t~+~\phi)}$
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The tangential velocity $\small{\omega\,A}$ is perpendicular to OP. So
the angle between the tangential velocity and the dotted line will be
$\small{\left[\frac{\pi}{2} - (\omega t + \phi) \right]}$
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So the horizontal component of the tangential velocity will be
$\small{\omega\,A \cos\left[\frac{\pi}{2} - (\omega t + \phi)
\right]\,=\,\omega\,A \sin (\omega t + \phi)}$
See identity 5 in the list of trigonometric identities.
•
But in the final expression, we must include a −ve sign because, the
horizontal component is acting towards the −ve side of the x-axis.
•
Thus we get: $\small{v(t)\,=\,-\omega\,A \sin (\omega t + \phi)}$
Note:
In
math classes, it is proved that, derivative of displacement, w.r.t time, will give velocity. This is true in the case of SHM also.
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{x(t)} &
{~=~} &{A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta} 2 } &{} &{v(t)} & {~=~} &{\frac{dx}{dt}}
\\
{~\color{magenta} 3 } &{\Rightarrow} &{v(t)}
& {~=~} &{\frac{d\left[A\cos(\omega t\,+\,\phi) \right]}{dt}}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{v(t)} & {~=~} &{-\omega A\sin(\omega t\,+\,\phi)}
\\ \end{array}}$
•
This is the same expression that we obtained above
Acceleration in simple harmonic motion
•
In fig.14.27 below, we consider the same magenta sphere again.
♦ Radius of the circle is A
♦ Angular speed of the sphere is $\small{\omega}$
 |
| Fig.14.27 |
•
We know that, the "projection of the sphere on the x-axis" will be in
SHM between (−A,0) and (A,0). We want an expression which can be used
to calculate the "acceleration of this projection" at any instant t.
This can be done in steps:
1. Let at any instant t, the position of the sphere be P.
We
know that, at any instant, the centripetal acceleration experienced by the sphere will
be $\small{v^2\,A}$ or $\small{\omega^2\,A}$. This is indicated by the white radial arrow
at P, in the fig.14.27 above.
2. Now we drop two perpendicular green dashed lines, onto the x-axis.
•
One through P and the other through the head of $\small{\omega\,A}$
•
These two perpendicular lines will give the projection of $\small{\omega^2\,A}$ on the x-axis.
3. This projection is the acceleration of the SHM at the instant t. It is denoted as $\small{a(t)}$. So our next aim is to find the expression for $\small{a(t)}$
4. Fig.14.28 below shows an enlarged view of the first quadrant.
 |
| Fig.14.28 |
•
The white dotted line is parallel to the x-axis and it passes through the head of $\small{\omega^2\,A}$. So the angle between $\small{\omega^2\,A}$ and the dotted line is $\small{(\omega t~+~\phi)}$
•
So the horizontal component of $\small{\omega^2\,A}$ will be $\small{\omega^2\,A \cos(\omega t + \phi)}$
•
But in the final expression, we must include a −ve sign because, the
horizontal component is acting towards the −ve side of the x-axis.
•
Thus we get: $\small{a(t)\,=\,-\omega^2\,A \cos (\omega t + \phi)}$
Note:
In
math classes, it is proved that, derivative of velocity, w.r.t time, will give acceleration. This is true in the case of SHM also.
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{v(t)} & {~=~} &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta} 2 } &{} &{a(t)} & {~=~} &{\frac{dv}{dt}}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{a(t)} & {~=~} &{\frac{d\left[-A \omega\sin(\omega t\,+\,\phi) \right]}{dt}}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{a(t)} & {~=~} &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$
• This is the same expression that we obtained above
Now we will see two important properties related to acceleration.
Property 1:
This can be explained in 4 steps:
1. We obtained: $\small{a(t)\,=\,-\omega^2\,A \cos (\omega t + \phi)}$
2. But in the previous section, we saw that: $\small{x(t)\,=\,A \cos (\omega t + \phi)}$
3. So we can write: $\small{a(t)\,=\,-\omega^2\,x(t)}$
4. Suppose that, $\small{\omega}$ is a constant. Then we can write:
acceleration is proportional to displacement.
Property 2:
This can be explained in steps:
1. In property 1, we saw that:
Acceleration = A constant × displacement
2. But there is a −ve sign in the expression for acceleration. So we can write:
♦ If the displacement is −ve, acceleration will be +ve.
♦ If the displacement is +ve, acceleration will be −ve.
Let us see how the two properties are applicable to the horizontal oscillating spring that we saw in fig.14.9 is section 14.1. It can be written as four cases.
Case 1
This can be written in 3 steps:
1. Consider the case when the mass is traveling from x=0 to x=A
In this case, the displacement will be +ve at any instant.
2. Since the displacement is +ve, acceleration will be −ve at any instant.
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That is, acceleration will be acting towards the −ve side of the x-axis
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That means, acceleration will be acting opposite to the direction of motion.
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That means, the mass will be experiencing deceleration.
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Indeed, due to this deceleration, the mass comes to a stop at x = A
3. Also, the deceleration experienced by the mass is not uniform.
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Greater the distance from the equilibrium position, greater will be the deceleration. That means, as the mass approaches x=A, it will be experiencing more and more deceleration.
Case 2
This can be written in 3 steps:
1. Consider the case when the mass is traveling from x=A to x=0
In this case, the displacement will be +ve at any instant.
2. Since the displacement is +ve, acceleration will be −ve at any instant.
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That is, acceleration will be acting towards the −ve side of the x-axis
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That means, acceleration will be acting in the direction of motion.
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Indeed, due to this acceleration, the mass will have the maximum possible velocity at x = 0
3. Also, the acceleration experienced by the mass is not uniform.
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Lesser the distance from the equilibrium position, lesser will be the acceleration. That means, as the mass approaches x = 0, it will be experiencing less and less acceleration.
Case 3
This is the case when the mass is traveling from x=0 to x=−A
The reader may write all the 3 steps in detail
Case 4
This is the case when the mass is traveling from x=−A to x=0
The reader may write all the 3 steps in detail
We have seen three functions:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x(t)} & {~=~} &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta} 2 } &{} &{v(t)} & {~=~} &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta} 3 } &{} &{a(t)} & {~=~} &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$
In the next section, we will see an analysis of their graphs
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