Two simple harmonic motions are represented by ${{y}_{1}}=5\sin (2\pi t+\dfrac{\pi }{6})$ and ${{y}_{2}}=5[\sin (3\pi t)+\sqrt{3}\cos (3\pi t)$.The ratio of their amplitude is
A.) 0.5
B.) 0.6
C.) 0.7
C.) 0.8
Answer
625.8k+ views
Hint:The Equation of simple harmonic motion is defined as $X=A\sin (\omega t)$ where (A) stands for amplitude and (t) stands for time period. The equation can be modified to different terms and can be written in different forms but this is the basic or fundamental form for an equation of a simple harmonic motion.
Complete Step by Step Solution:
We have been given with two simple harmonic equations of motion
${{y}_{1}}=5\sin (2\pi t+\dfrac{\pi }{6})$
${{y}_{2}}=5[\sin (3\pi t)+\sqrt{3}\cos (3\pi t)]$
And we know that the basic way to write the equation of motion is $X=A\sin (\omega t)$ where (A) stands for amplitude and (t) stands for time period.
Now on comparing equation ${{y}_{1}}$ with the basic equation of motion we get
\[5\sin (2\pi t+\dfrac{\pi }{6})\] $=A\sin (\omega t)$
Now from the above equation we got to know that A= 5 because the equation is already in its fundamental form
But the equation ${{y}_{2}}$ is not in its fundamental form and hence we need to make it in this basic form
We have ${{y}_{2}}=5[\sin (3\pi t)+\sqrt{3}\cos (3\pi t)]$
on solving the equation, we get,
${{y}_{2}}=5\sin (3\pi t)+5\sqrt{3}\cos (3\pi t)$
So now to find the magnitude of amplitude of such equations we have a formula A=$\sqrt{{{({{A}_{1}})}^{2}}+{{({{A}_{2}})}^{2}}}$
Where ${{A}_{1}}=5$ and ${{A}_{2}}=5\sqrt{3}$
On purring the values in the formula, we get
A=$\sqrt{{{(5)}^{2}}+{{(5\sqrt{3})}^{2}}}$
A= 10
So, the amplitude for ${{y}_{2}}$is 10 units.
Now according to the question, we need to find the ratio of amplitude of ${{y}_{1}}$with ${{y}_{2}}$.
$\dfrac{{{y}_{1}}}{{{y}_{2}}}=\dfrac{5}{10}$
So, $\dfrac{{{y}_{1}}}{{{y}_{2}}}=0.5$
Hence, we can conclude that option (A) is the correct answer.
Note:
When we need to find the amplitude of the equation of the simple harmonic motion, always try to bring the equation of simple harmonic motion in its original form. It will make it easy to compare the amplitude to its original form. And do not forget to find the magnitude of amplitude if two amplitudes are given in one equation of a simple harmonic motion.
Complete Step by Step Solution:
We have been given with two simple harmonic equations of motion
${{y}_{1}}=5\sin (2\pi t+\dfrac{\pi }{6})$
${{y}_{2}}=5[\sin (3\pi t)+\sqrt{3}\cos (3\pi t)]$
And we know that the basic way to write the equation of motion is $X=A\sin (\omega t)$ where (A) stands for amplitude and (t) stands for time period.
Now on comparing equation ${{y}_{1}}$ with the basic equation of motion we get
\[5\sin (2\pi t+\dfrac{\pi }{6})\] $=A\sin (\omega t)$
Now from the above equation we got to know that A= 5 because the equation is already in its fundamental form
But the equation ${{y}_{2}}$ is not in its fundamental form and hence we need to make it in this basic form
We have ${{y}_{2}}=5[\sin (3\pi t)+\sqrt{3}\cos (3\pi t)]$
on solving the equation, we get,
${{y}_{2}}=5\sin (3\pi t)+5\sqrt{3}\cos (3\pi t)$
So now to find the magnitude of amplitude of such equations we have a formula A=$\sqrt{{{({{A}_{1}})}^{2}}+{{({{A}_{2}})}^{2}}}$
Where ${{A}_{1}}=5$ and ${{A}_{2}}=5\sqrt{3}$
On purring the values in the formula, we get
A=$\sqrt{{{(5)}^{2}}+{{(5\sqrt{3})}^{2}}}$
A= 10
So, the amplitude for ${{y}_{2}}$is 10 units.
Now according to the question, we need to find the ratio of amplitude of ${{y}_{1}}$with ${{y}_{2}}$.
$\dfrac{{{y}_{1}}}{{{y}_{2}}}=\dfrac{5}{10}$
So, $\dfrac{{{y}_{1}}}{{{y}_{2}}}=0.5$
Hence, we can conclude that option (A) is the correct answer.
Note:
When we need to find the amplitude of the equation of the simple harmonic motion, always try to bring the equation of simple harmonic motion in its original form. It will make it easy to compare the amplitude to its original form. And do not forget to find the magnitude of amplitude if two amplitudes are given in one equation of a simple harmonic motion.
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