Superposition of Harmonic Oscillations
Solved Problems
Example 1.
Write the expression in the form \(A\sin \left( {\omega t + \alpha } \right):\) \[12\sin 2t + 5\cos 2t.\]
Solution.
The oscillatory process has the angular frequency \(\omega = 2\) and is described by the function
\[y\left( t \right) = 12\sin 2t + 5\cos 2t = {C_1}\sin 2t + {C_2}\sin 2t.\]
Calculate the amplitude of oscillations:
\[A = \sqrt {C_1^2 + C_2^2} = \sqrt {{{12}^2} + {5^2}} = \sqrt {144 + 25} = \sqrt {169} = 13.\]
The initial phase \(\alpha\) is defined by the formulas:
\[\cos \alpha = \frac{{{C_1}}}{A} = \frac{{12}}{{13}},\;\sin \alpha = \frac{{{C_2}}}{A} = \frac{5}{{13}}.\]
Since sine and cosine are positive, the angle \(\alpha\) is in the first quadrant. The value of \(\alpha\) is given by
\[\alpha = \arccos \frac{{12}}{{13}}.\]
Hence,
\[y\left( t \right) = 12\sin 2t + 5\cos 2t = 13\sin \left( {2t + \alpha } \right),\]
where \(\alpha = \arccos \frac{{12}}{{13}}.\)
Example 2.
Write the expression in the form \(A\sin \left( {\omega t + \alpha } \right):\) \[21\sin 3t - 20\cos 3t.\]
Solution.
The function has the angular frequency \(\omega = 3.\) Denoting \({C_1} = 21\) and \({C_2} = -20,\) find the amplitude of oscillations:
\[A = \sqrt {C_1^2 + C_2^2} = \sqrt {{{21}^2} + {{\left( { - 20} \right)}^2}} = \sqrt {441 + 400} = \sqrt {841} = 29.\]
We introduce the angle \(\alpha\) such that
\[\cos \alpha = \frac{{{C_1}}}{A} = \frac{{21}}{{29}},\;\sin \alpha = \frac{{{C_2}}}{A} = - \frac{{20}}{{29}}.\]
We see that \(\cos \alpha \gt 0\) and \(\sin \alpha \lt 0,\) so the angle \(\alpha\) lies in the \(4\text{th}\) quadrant. It can be expressed as
\[\alpha = \arcsin \left( { - \frac{{20}}{{29}}} \right) = - \arcsin \frac{{20}}{{29}}.\]
Therefore,
\[y\left( t \right) = 21\sin 3t - 20\cos 3t = 29\sin \left( {3t + \alpha } \right),\]
where \(\alpha = - \arcsin \frac{{20}}{{29}}.\)
Example 3.
Find the greatest and least value of the function \[y\left( t \right) = 8\sin t - 15\cos t.\]
Solution.
The function \(y\left( t \right)\) can be written in the form
\[y\left( t \right) = A\sin \left( {\omega t + \alpha } \right).\]
So, to determine the greatest and least value, we need to calculate the amplitude \(A.\) Let \({C_1} = 8,\) \({C_2} = -15.\) Then
\[A = \sqrt {C_1^2 + C_2^2} = \sqrt {{8^2} + {{\left( { - 15} \right)}^2}} = \sqrt {64 + 225} = \sqrt {289} = 17.\]
The range of the sine function is \(\left[ { - 1,1} \right].\) Hence, the greatest value of \(y\left( t \right)\) is \(17,\) and the least value is \(-17.\)
Example 4.
Find the greatest and least value of the function \[y\left( t \right) = 24\sin \frac{t}{2} + 7\cos \frac{t}{2}.\]
Solution.
This function can be converted to the form \(y\left( t \right) = A\sin \left( {\omega t + \alpha } \right),\) where its greatest and least values are determined by the amplitude \(A.\) Let's denote \({C_1} = 24\) and \({C_2} = 7.\) Then
\[A = \sqrt {C_1^2 + C_2^2} = \sqrt {{{24}^2} + {7^2}} = \sqrt {576 + 49} = \sqrt {625} = 25.\]
Thus, the greatest value is \(25,\) and the least value is \(-25.\)
Example 5.
Find the sum of harmonic oscillations:
\[{y_1}\left( t \right) = 2\sin t,\;{y_2}\left( t \right) = 2\cos \left( {t + \frac{\pi }{6}} \right).\]
Solution.
Using the cosine addition formula, we rewrite the function \({y_2}\left( t \right)\) as follows:
\[{y_2}\left( t \right) = 2\cos \left( {t + \frac{\pi }{6}} \right) = 2\left( {\cos t\cos \frac{\pi }{6} - \sin t\sin \frac{\pi }{6}} \right) = 2\left( {\frac{{\sqrt 3 }}{2}\cos t - \frac{1}{2}\sin t} \right) = \sqrt 3 \cos t - \sin t.\]
Then the sum of two oscillations is given by
\[y\left( t \right) = {y_1}\left( t \right) + {y_2}\left( t \right) = 2\sin t + \sqrt 3 \cos t - \sin t = \sin t + \sqrt 3 \cos t.\]
We can represent the result in the form \(A\sin \left( {\omega t + \alpha } \right):\)
\[y\left( t \right) = \sin t + \sqrt 3 \cos t = 2\left( {\frac{1}{2}\sin t + \frac{{\sqrt 3 }}{2}\cos t} \right) = 2\left( {\cos \frac{\pi }{3}\sin t + \sin \frac{\pi }{3}\cos t} \right) = 2\sin \left( {t + \frac{\pi }{3}} \right).\]
Example 6.
Find the sum of harmonic oscillations:
\[{y_1}\left( t \right) = \sqrt 2 \sin 8t,\;{y_2}\left( t \right) = \sqrt 2 \cos 8t.\]
Solution.
Using the sine addition identity, we represent the sum of the functions \({y_1}\left( t \right)\) and \({y_2}\left( t \right)\) in the form \(A\sin \left( {\omega t + \alpha } \right):\)
\[y\left( t \right) = {y_1}\left( t \right) + {y_2}\left( t \right) = \sqrt 2 \sin 8t + \sqrt 2 \cos 8t = 2\left( {\frac{{\sqrt 2 }}{2}\sin 8t + \frac{{\sqrt 2 }}{2}\cos 8t} \right) = 2\left( {\cos \frac{\pi }{4}\sin 8t + \sin \frac{\pi }{4}\cos 8t} \right) = 2\sin \left( {8t + \frac{\pi }{4}} \right).\]
Example 7.
Find the sum of harmonic oscillations and determine the beat frequency:
\[{y_1}\left( t \right) = \sin \left( {40\pi t} \right),\;{y_2}\left( t \right) = \sin \left( {42\pi t} \right).\]
Solution.
By the sum-to-product identity, we have
\[y\left( t \right) = {y_1}\left( t \right) + {y_2}\left( t \right) = \sin \left( {40\pi t} \right) + \sin \left( {42\pi t} \right) = 2\sin \left( {\frac{{40 + 42}}{2}\pi t} \right)\cos \left( {\frac{{40 - 42}}{2}\pi t} \right) = 2\sin \left( {41\pi t} \right)\cos \left( { - \pi t} \right) = 2\cos \left( {\pi t} \right)\sin \left( {41\pi t} \right).\]
Find the frequencies of oscillations \({f_1}\) and \({f_2}:\)
\[40\pi t = 2\pi {f_1}t, \;\Rightarrow 2{f_1} = 40, \;\Rightarrow {f_1} = 20\,Hz;\]
\[42\pi t = 2\pi {f_2}t, \;\Rightarrow 2{f_2} = 42, \;\Rightarrow {f_2} = 21\,Hz;\]
Hence, the beat frequency is equal to
\[{f_{\text{beat}}} = \left| {{f_1} - {f_2}} \right| = \left| {20 - 21} \right| = 1\,Hz.\]
