# Superposition of Harmonic Oscillations

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Write the expression in the form Asin(ωt + α): $12\sin 2t + 5\cos 2t.$

### Example 2

Write the expression in the form Asin(ωt + α): $21\sin 3t - 20\cos 3t.$

### Example 3

Find the greatest and least value of the function $y\left( t \right) = 8\sin t - 15\cos t.$

### Example 4

Find the greatest and least value of the function $y\left( t \right) = 24\sin \frac{t}{2} + 7\cos \frac{t}{2}.$

### Example 5

Find the sum of harmonic oscillations:

${y_1}\left( t \right) = 2\sin t,\;{y_2}\left( t \right) = 2\cos \left( {t + \frac{\pi }{6}} \right).$

### Example 6

Find the sum of harmonic oscillations:

${y_1}\left( t \right) = \sqrt 2 \sin 8t,\;{y_2}\left( t \right) = \sqrt 2 \cos 8t.$

### Example 7

Find the sum of harmonic oscillations and determine the beat frequency:

${y_1}\left( t \right) = \sin \left( {40\pi t} \right),\;{y_2}\left( t \right) = \sin \left( {42\pi t} \right).$

### 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

${f_{\text{beat}}} = \left| {{f_1} - {f_2}} \right| = \left| {20 - 21} \right| = 1\,Hz.$