In the previous section, we saw that displacement from the equilibrium position can be specified using sine or cosine functions. In this section, we will see simple harmonic motion.
1. Consider a particle subjected to oscillation along the x-axis. Suppose that, the oscillation satisfies the following conditions:
(i) -`A and A are the extreme points.
(ii) The origin O is the equilibrium point.
(iii) Displacement from the origin can be specified using a sinusoidal function.
(Sine and cosine functions are examples for sinusoidal functions)
2. We have seen that the particle attached to a horizontal spring can satisfy the above three conditions. (see section 14.1)
Let us observe this oscillation. Assume that, the stop-watch is turned on when the particle is at O
3. Then, in addition to those three conditions, the particle satisfies more conditions:
(iv) When $\small{t~=~0}$, the particle is at O, and the velocity is maximum.
(v) When $\small{t~=~\frac{T}{4}}$, the particle is at +A, and the velocity is zero.
(vi) When $\small{t~=~\frac{T}{2}}$, the particle is at O, and the velocity is maximum.
(vii) When $\small{t~=~\frac{3T}{4}}$, the particle is at -`A, and the velocity is zero.
(viii) When $\small{t~=~T}$, the particle is at O, and the velocity is maximum.
(ix) When $\small{t~=~\frac{5T}{4}}$, the particle is at +A, and the velocity is zero.
so on . . .
4. If the nine conditions are satisfied, then that oscillation is called a simple harmonic motion. It is abbreviated as SHM.
5. In the previous sections, we became familiar with the function:
$\small{x(t)~=~A \cos\left(\omega t \,+\,\phi \right)}$
♦ $\small{x(t)}$ is the displacement $\small{x}$ as a function of time $\small{t}$
♦ $\small{A}$ is the amplitude
♦ $\small{\omega}$ is the angular frequency
♦ $\small{\omega t \,+\,\phi}$ is the phase. It depends on time.
♦ $\small{\phi}$ is the phase constant
• This is a sinusoidal function which can represent a SHM.
Let us see some examples:
Example 1:
In fig.14.21 below, the red and green curves represent two independent simple harmonic motions.
 |
| Fig.14.21 |
From the graph, we get two information:
(i) Information about period T
• Both red and green reach maximums at the same instants.
• Both red and green reach minimums at the same instants
• Both red and green reach zero at the same instants
• So red and green have the same phase $\small{\left(\omega t \,+\,\phi \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
Green: $\small{x(t)~=~3 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
The instants corresponding to maximums of red, can be obtained by solving the equation:
$\small{4~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
The instants corresponding to minimums of red, can be obtained by solving the equation:
$\small{-4~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
The instants corresponding to zeros of red, can be obtained by solving the equation:
$\small{0~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
The same procedure can be used for green.
Always remember the relation between T and $\small{\omega}$, which is: $\small{\omega~=~\frac{2 \pi}{T}}$
So for the red, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\omega t} & {~=~} &{\frac{\pi t}{6}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{\omega} & {~=~} &{\frac{\pi}{6}~=~\frac{2 \pi}{T}} \\
{~\color{magenta} 3 } &{\Rightarrow} &{T} & {~=~} &{12~\text{seconds}} \\
\end{array}}$
In the same way, we can find the period of green also.
(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of red is 4 units.
• The two magenta horizontal dashed lines indicate that, the amplitude of green is 3 units.
◼ Based on this example, we can write:
Two simple harmonic motions may have the same phase $\small{\left(\omega t \,+\,\phi \right)}$. But they can have different amplitudes.
Example 2:
In fig.14.22 below, the red and green curves represent two independent simple harmonic motions.
 |
| Fig.14.22 |
From the graph, we get two information:
(i) Information about period T
• Red and green do not reach maximums at the same instants.
• Red and green do not reach minimums at the same instants
• Red and green do not reach zero at the same instants
• So red and green do not have the same phase $\small{\left(\omega t \,+\,\phi \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \right)}$
Green: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \,-\,\frac{\pi}{4} \right)}$
(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of both red and green is 4 units.
◼ Based on this example, we can write:
Two
simple harmonic motions may have the same amplitude
$\small{A}$. But they can have different phases.
Example 3:
In fig.14.23 below, the red and green curves represent two independent simple harmonic motions.
 |
| Fig.14.23 |
From the graph, we get two information:
(i) Information about period T
• Consider the magenta vertical dashed line. Both red and green reach maximum at the time indicated by this dashed line.
• The "horizontal distance between y-axis and the magenta vertical dashed line" indicates a time duration. Within this time duration, green completes two cycles. Red completes only one cycle.
• So it is clear that, $\small{T_{\text{green}}}$ is half of $\small{T_{\text{red}}}$
• So red and green do not have the same angular frequency $\small{\left(\omega \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{4} \right)}$
Green: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{2} \,-\,\frac{\pi}{4} \right)}$
(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of both red and green is 4 units.
◼ Based on this example, we can write:
Two
simple harmonic motions may have the same amplitude
$\small{A}$. But they can have different angular frequencies.
Now we will see a solved example:
Solved example 14.4
Which of the following function of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic ? Give the period for each case
(i) $\small{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}$
(ii) $\small{\sin^2 \left(\omega t \right)}$
Solution:
Part (i):
1. We are given: $\small{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}$
• Let us try to rearrange the expression into the standard form:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)} & {~=~} &{\sin\left(\omega t \right)~-~\sin\left(\frac{\pi}{2} - \omega t \right)} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{2 \cos\left(\frac{\pi}{4} \right)\sin\left(\omega t - \frac{\pi}{4} \right)} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{2 \left(\frac{1}{\sqrt 2} \right)\sin\left(\omega t - \frac{\pi}{4} \right)} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\sqrt 2\,\sin\left(\omega t - \frac{\pi}{4} \right)} \\
\end{array}}$
◼ Remarks:
• 1 (magenta color):
Here we use the identity 6 in the list of trigonometric identities.
• 2 (magenta color):
Here we use the identity 20(d) in the list of trigonometric identities.
2. Based on this standard form, we can write:
The given expression represents a SHM with,
♦ Amplitude $\small{\sqrt 2}$ units
♦ Angular frequency $\small{\omega}$
♦ Phase constant $\small{- \frac{\pi}{4}}$
3. The standard form obtained in (1) can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x(t)} & {~=~} &{\sqrt 2\,\sin\left(\omega t - \frac{\pi}{4} \right)} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt 2\,\sin\left[2\pi~+~\left(\omega t - \frac{\pi}{4} \right) \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\sqrt 2\,\sin\left(\omega t + \frac{7\pi}{4} \right)} \\
\end{array}}$
• So for the given SHM, the phase constant $\small{- \frac{\pi}{4}}$ is same as $\small{\frac{7\pi}{4}}$.
Part (ii):
1. We are given: $\small{\sin^2 \left(\omega t \right)}$
• Let us try to rearrange the expression into the standard form:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\sin^2 \left(\omega t \right)} & {~=~} &{\frac{1\,-\,\cos\left(2\omega t \right)}{2}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{1}{2}~-~\frac{\cos\left(2\omega t \right)}{2}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{1}{2}~-~\frac{1}{2}\,\cos\left(2\omega t \right)} \\
\end{array}}$
◼ Remarks:
• 1 (magenta color):
Here we use the identity 14 in the list of trigonometric identities.
2. In fig.14.24 below, two graphs are shown.
 |
| Fig.14.24 |
• Green represents: $\small{x(t)~=~0.5 \cos\left( 2\omega t \right)}$
♦ The two horizontal magenta dashed lines indicate that, the amplitude of green is 0.5 units.
♦ This graph is familiar to us. The equilibrium point is at O.
• Red represents: $\small{x(t)~=~0.5\,-\,0.5 \cos\left( 2\omega t \right)}$
♦ This is also a sinusoidal curve.
♦ The horizontal yellow dashed line and the x-axis indicate that, the distance between extreme points is 1 unit.
♦ So the amplitude of red is 1/2, which is 0.5 units.
♦ The upper horizontal magenta dashed line divides the red into two equal upper and lower parts. This dashed line passes through 0.5 on the y-axis. That means, equilibrium point of red is 0.5 units away from the origin.
3. Comparing the green with the standard form, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{2\omega} & {~=~} &{\frac{2\pi}{T}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{\omega} & {~=~} &{\frac{\pi}{T}} \\
{~\color{magenta} 3 } &{\Rightarrow} &{T} & {~=~} &{\frac{\pi}{\omega}} \\
\end{array}}$
In the next
section, we will see velocity and acceleration in simple harmonic motion.
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