Question

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

Answer 1

Hint
First we need to simplify the wave equation that is given in the question to the form of $y = A\sin \left( {\omega t + \phi } \right)$ . Next we need to compare our equation to the general form of wave equation and from there we will get the amplitude and the time period.
Formula Used: In this solution we will be using the following formula,
$\Rightarrow y = A\sin \left( {\omega t + \phi } \right)$
where $A$ is the amplitude,
$\omega$ is the frequency and $t$ is the time
and $\phi$ is the phase difference.
and $\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$
where $A,B$ are angles.

Complete step by step answer
In the question we are given that the displacement of a wave is given by the formula, $y = 5\left( {\sin 3\pi t + \sqrt 3 \cos 3\pi t} \right)cm$
Now we need to first simplify this to a more general form. So we multiply 2 in the numerator and the denominator.
So we have
$\Rightarrow y = 2 \times 5\left( {\dfrac{1}{2}\sin 3\pi t + \dfrac{{\sqrt 3 }}{2}\cos 3\pi t} \right)cm$
In this above equation, we can write the $\dfrac{1}{2}$ as $\cos \dfrac{\pi }{3}$ as $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ and we can write the $\dfrac{{\sqrt 3 }}{2}$ as $\sin \dfrac{\pi }{3}$ as, $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ . So we get by substituting,
$\Rightarrow y = 2 \times 5\left( {\cos \dfrac{\pi }{3}\sin 3\pi t + \sin \dfrac{\pi }{3}\cos 3\pi t} \right)cm$
Now the formula, $\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$
So here we take, $A = 3\pi t$ and $B = \dfrac{\pi }{3}$
So we get in the equation of motion,
$\Rightarrow y = 10\sin \left( {3\pi t + \dfrac{\pi }{3}} \right)cm$
Now we can compare this with the general wave equation, $y = A\sin \left( {\omega t + \phi } \right)$ we get,
$A = 10cm$ So this is the amplitude of the motion.
and $\omega = 3\pi$ . Here $\omega$ is the frequency and we can get the time period from the frequency by the formula,
$\Rightarrow T = \dfrac{{2\pi }}{\omega }$ . So substituting we get,
$\Rightarrow T = \dfrac{{2\pi }}{{3\pi }}$
Cancelling the $\pi$ we get the time period as, $T = \dfrac{2}{3}s$
Hence the amplitude is $10cm$ and time period is $\dfrac{2}{3}s$ . So the correct option is (D).

Note
This given equation is the equation of a simple harmonic motion. This is a special kind of motion in which the restoring force of the moving object is directly proportional to the magnitude of the displacement of the object. The direction of this restoring force will be towards the equilibrium position of the object.