In the previous section, we saw period and frequency. In this section, we will see displacement.
Let us first see the details of an oscillation. It can be written in 6 steps:
1.
In fig.14.3 (a) below, a red block of mass m, is attached to a spring.
The other end of the spring is fixed to a rigid wall. The block is
resting on a friction-less surface.
 |
| Fig.14.9 |
2. In fig.14.3 (b), the block is pulled towards the right by a distance A. From this position, the block is released from rest. It will then travel towards the left. Even after traveling a distance 'A', it will continue to travel towards the left. That is., even after passing the point x= 0, it will continue to travel towards the left. This is shown in fig.c.
3. But once the point x = 0 is passed, the block will begin to experience a resistive force. The block is able to overcome the resistive force, and reach up to x = −A. Once it reaches x=−A, it stops. That is., it's velocity becomes zero. This is shown in fig.d.
4. Then it starts the reverse journey. In the reverse journey, even after passing the point x=0, it will continue to travel towards the right. This is shown in fig.e.
5. But once the point x = 0 is passed, the block will begin to experience a resistive force. The block is able to overcome the resistive force, and reach up to x = A. Once it reaches x=A, it stops. That is., it's velocity becomes zero. This is shown in fig.f.
6. At this point, one cycle is complete. Then it again starts the journey towards left. This process continues giving rise to continuous oscillation.
Displacement
The
above 6 steps give us a basic understanding about oscillation. Now we
will derive an expression for displacement. It can be done in 9 steps:
1.
We saw that, the red block is oscillating between x=A and x=−A. Let us
mark those two points as P and Q respectively, on the x-axis. This is
shown in fig.14.4 below:
 |
| Fig.14.4 |
• The coordinates of P and Q are (A,0) and (−A,0) respectively. So O is the equilibrium position of the block.
2. Draw the red circle with center at O and radius equal to A.
• The magenta sphere is performing uniform circular motion along the red circle.
•
The angular velocity of the magenta sphere is $\small{\omega}$.
•
That means, the angle subtended by the sphere, at O, in each second, is $\small{\omega}$.
3.
Consider the instant when the magenta sphere is at M. At that instant,
the line OM makes an angle $\small{\theta}$ with the x-axis.
•
Drop the perpendicular MN from M, onto the x-axis.
•
From the right triangle OMN, we get:
$\small{ON~=~ OM\,\cos \theta~=~A\,\cos\theta}$.
4. Now, ON is the horizontal displacement of the magenta sphere from the equilibrium position O. So we get a method to write the horizontal displacement of the magenta sphere from O.
◼ Let us check for other points. For that, we will try to write a general method which is applicable to all points.
(i) In fig.14.5(a) below, M is in the I quadrant.
♦ On OM, Mark M' such that, OM' = 1 unit.
♦ Drop the perpendicular M'N' onto the x-axis.
 |
| Fig.14.5 |
• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM}{OM'}~=~\frac{ON}{ON'}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the x-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the x-coordinate of M' is $\small{\cos \theta}$ (Details here)
• So we get:
$\small{ON~=~A \left(\frac{\cos \theta}{1} \right)~=~A\,\cos \theta}$
(ii) In fig.14.5(b) above, M is in the II quadrant.
♦ On OM, Mark M' such that, OM' = 1 unit.
♦ Drop the perpendicular M'N' onto the x-axis.
• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM}{OM'}~=~\frac{ON}{ON'}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the x-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the x-coordinate of M' is $\small{\cos \theta}$
• So we get:
$\small{ON~=~A \left(\frac{\cos \theta}{1} \right)~=~A\,\cos \theta}$
•
Since M is in the II quadrant, $\small{\cos \theta}$ will be −ve.
Indeed, N will have a −ve x-coordinate because it is on the −ve side of
the x-axis.
(iii) In fig.14.5(c) above, M is in the III quadrant.
We can write similar steps and obtain the same result.
(iv) In fig.14.5(d) above, M is in the IV quadrant.
We can write similar steps and obtain the same result.
5.
We see that, this method is very effective to write the horizontal
displacement of the magenta sphere. But we want the horizontal
displacement of the red block.
•
So we assume that:
♦ At the instant when the block is released from P, the sphere starts the revolution from P
♦ At the instant when the block reaches O, the sphere reaches R
♦ At the instant when the block reaches Q, the sphere also reaches Q
♦ At the instant when the block returns through O, the sphere reaches S
♦ At the instant when the block returns back at P, the sphere also reaches back at P
6.
That means, the time period (T) for one oscillation of the block is
same as the time for one revolution of the sphere. Using this
information, we can write an expression for $\small{\theta}$
•
Time for one revolution of the sphere = T seconds
⇒ Time for $\small{2 \pi}$ radians = T seconds
⇒ Time for 1 radian = $\small{\frac{T}{2 \pi}}$ seconds
⇒ Angular distance covered in 1 second = $\small{\frac{2 \pi}{T}}$ radian
•
The stop-watch is turned on at the instant when the block is released
from P. At that same instant, the sphere starts the revolution from the
same point P.
•
Consider the instant at which the reading in the stop-watch is t. Let at that instant, the sphere be at M.
•
So when the reading is t, the angular distance covered is $\small{\theta}$
•
Angular distance covered in 1 second = $\small{\frac{2 \pi}{T}}$ radian
⇒ Angular distance covered in t seconds = $\small{\frac{2 \pi t}{T}}$ radian
⇒ $\small{\theta}$ = $\small{\frac{2 \pi t}{T}}$ radian
•
So we can write:
At any time t,
The horizontal displacement of the sphere from O
= The horizontal displacement of the block from O
= $\small{A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)}$
•
That means:
$\small{x(t)~=~A\,\cos\left(\frac{2 \pi t}{T} \right)}$
7. We derived the equation: $\small{x(t)~=~A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)}$.
•
If we want, we can use $\small{\omega}$ instead of $\small{T}$.
•
We have:
the angle subtended by the sphere, at O, in each second, is $\small{\omega}$.
•
So in $\small{t}$ seconds, the sphere will subtend $\small{\omega\,t}$ radians at O
•
But the angle subtended by the sphere, at O, in $\small{t}$ seconds, is $\small{\theta}$.
•
That means: $\small{\theta~=~\omega\,t}$
•
Thus we get:
$\small{x(t)~=~A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)~=~A\,\cos\left(\omega\,t \right)}$.
8. Here, the only variable on the R.H.S is $\small{t}$.
♦ $\small{t}$ is the independent variable
♦ $\small{x}$ is the dependent variable
•
So while drawing the graph, we must plot $\small{t}$ along the x-axis and $\small{x}$ along the y-axis.
9. One such graph is shown in fig.14.6 below.
♦ A is assumed to be 2.5 units
♦ T is assumed to be 6 s
 |
| Fig.14.6 |
•
We get:
$\small{x(t)~=~A\,\cos\left(\frac{2 \pi t}{T}
\right)~=~(2.5)\,\cos\left(\frac{2 \pi t}{6}
\right)~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
•
Let us write some of the information that can be obtained from the graph:
(i) To find the instants at which the red block reaches the positive extreme, we need to solve the equation:
$\small{x(t)~=~2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
$\small{\Rightarrow 1~=~\cos\left(\frac{\pi t}{3} \right)}$
•
This is a trigonometrical equation. We have seen the method to solve such equations, in our math classes (Details here).
•
We get: t = 0, 6, 12, . . .
•
The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all 2.5
(ii) To find the instants at which the red block reaches the negative extreme, we need to solve the equation:
$\small{x(t)~=~-2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
$\small{\Rightarrow -1~=~\cos\left(\frac{\pi t}{3} \right)}$
•
This is a trigonometrical equation. Solving it, we get:
t = 3, 9, 6, . . .
•
The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all −2.5
(iii) To find the instants at which the red block reaches the equilibrium position, we need to solve the equation:
$\small{x(t)~=~0~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
$\small{\Rightarrow 0~=~\cos\left(\frac{\pi t}{3} \right)}$
•
This is a trigonometrical equation. Solving it, we get:
t = 1.5, 4.5, 7.5, 10.5, . . .
•
The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all zero.
Phase constant
This can be explained in 3 steps:
1.
Consider fig.14.4 again. We turned the stop-watch on, when the red
sphere was at P. Then we noted the time 't' at which the red sphere is
at the arbitrary point M. The angle MOP at time 't' is denoted as
'$\small{\theta}$'
2. The same fig.14.4 is modified and shown again in fig.14.7 below:
 |
| Fig.14.7 |
• Here, the stop-watch is turned on, when the sphere is at B. At B, the line OB already makes an angle $\small{\phi}$ with the x-axis.
• So at time 't', the sphere is at the arbitrary point M and the line OM makes an angle of $\small{\theta + \phi}$ with the x-axis
• Therefore in this case, the horizontal displacement of the sphere from O, at time 't' is given by:
$\small{x(t)~=~A \,\cos\left(\theta + \phi \right)}$
3.
Let us draw the graph. It is shown in fig.14.8 below. For easy
comparison, the previous graph in red color, is also shown as such. The
new graph is shown in green color. For the new graph:
♦ A is the same 2.5 units
♦ T is the same 6 s
♦ $\small{\phi}$ is assumed to be $\small{\frac{\pi}{6}}$
 |
| Fig.14.8 |
•
We get:
$\small{x(t)~=~A\,\cos\left(\frac{2 \pi t}{T}+\phi
\right)~=~(2.5)\,\cos\left(\frac{2 \pi t}{6}+\frac{\pi}{6}
\right)~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
•
Let us write some of the information that can be obtained from the graph:
(i) To find the instants at which the red block reaches the positive extreme, we need to solve the equation:
$\small{x(t)~=~2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
$\small{\Rightarrow 1~=~\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
•
This is a trigonometrical equation. Solving it, we get:
t = 5.5, 11.5, . . .
•
The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all 2.5
(ii) To find the instants at which the red block reaches the negative extreme, we need to solve the equation:
$\small{x(t)~=~-2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
$\small{\Rightarrow -1~=~\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
•
This is a trigonometrical equation. Solving it, we get:
t = 2.5, 8.5, . . .
•
The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all −2.5
(iii) To find the instants at which the red block reaches the equilibrium point, we need to solve the equation:
$\small{x(t)~=~0~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
$\small{\Rightarrow 0~=~\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
•
This is a trigonometrical equation. Solving it, we get:
t = 1, 4, 7, 10, . . .
•
The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all zero.
(iv) The green curve is an exact replica of the red curve. But the green is shifted by a small amount towards the left.
•
That means:
The green reaches the extreme points and zero points earlier than the red.
•
For example:
♦ Red reaches a maximum at t = 6 s. Green reaches a maximum at t = 5.5 s
♦ Red reaches a minimum at t = 3 s. Green reaches a minimum at t = 2.5 s
♦ Red reaches a zero at t = 7.5 s. Green reaches a maximum at t = 7 s
•
We say that:
The two oscillations are out of phase by an angle of $\small{\frac{\pi}{6}}$ radians
•
The angle $\small{\phi~=~\frac{\pi}{6}}$
•
$\small{\phi}$ is called the phase constant.
In the next
section, we will see the oscillation of the same spring in the vertical direction.
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