Question

A simple harmonic motion is represented by ${\text{y }} = {\text{ 5(}}sin{\text{ }}\left( {3\pi t} \right) + \sqrt 3 \cos (3\pi t))cm$. The amplitude and time period of the motion are
A) $10cm,\dfrac{2}{3}s.$
B) $5cm,\dfrac{2}{3}s.$
C) $10cm,\dfrac{3}{2}s.$
D) $5cm,\dfrac{3}{2}s.$

Answer 1

For such questions, First compare the given equation with the general equation of the simple harmonic motion (SHM). If the given equation is not in general form then manipulate the given equation with some mathematical changes to make it comparable with the general equation. Now compare the two equations and then find each value like amplitude,(angular velocity) $\omega$ of the given object.

Complete step by step answer:
For comparison, we must know the general equation of SHM.
The general equation of the SHM is
$x{\text{ }} = {\text{ }}A{\text{ }}sin{\text{ }}\left( {\omega t{\text{ }} + {\text{ }}\Phi } \right)$
Where, $x$ is the position of the object, $A$ is amplitude.
$\left( {\omega t{\text{ }} + {\text{ }}\Phi } \right)$ is used to determine the phase angle of the object and $\omega$ is angular velocity.
We also know that time period is
$T = \dfrac{{2\pi }}{\omega }$
Now, our given equation is $y = 5(sin3\pi t + \sqrt 3 cos3\pi t)$
But it’s not matching with the general equation of SHM so what we do is we try to manipulate it with some trigonometric quantities,
Now, from the question, we will try to convert given eq. in the general form
$\Rightarrow y = 5(sin3\pi t + \sqrt 3 cos3\pi t)$
To make it comparable, we first multiply and divide it with 2.
$\Rightarrow y = 5 \times 2(\dfrac{1}{2} \times sin3\pi t + \dfrac{{\sqrt 3 }}{2}cos3\pi t)$
We know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow y = 5 \times 2(\cos \dfrac{\pi }{3} \times sin3\pi t + \sin \dfrac{\pi }{3}cos3\pi t)$
Now clearly we can see the formula of $\sin (A + B)=\sin A.\cos B+\cos A. \sin B$ in the above equation
$\Rightarrow y = 10(\sin (3\pi t - \dfrac{\pi }{3}))$
Comparing it with the general equation we get
$\Rightarrow A = 10cm$ and $\omega = 3\pi$
$\Rightarrow T = \dfrac{{2\pi }}{\omega }$
$\Rightarrow T= \dfrac{{2\pi }}{{3\pi }}$
$\Rightarrow T= \dfrac{2}{3}s$

The amplitude of the motion is 10cm and the time period of the motion is $\dfrac{2}{3}s$. Hence, option (A) is correct.

Note:
Linear Simple Harmonic Motion: When a particle moves to and fro about a fixed point (called equilibrium position) along with a straight line then its motion is called linear Simple Harmonic Motion.
Angular Simple Harmonic Motion: When a system oscillates angular long with respect to a fixed axis then its motion is called angular simple harmonic motion.