Question

A particle is subjected to two simple harmonic motions given by ${x_1} = 2 \cdot 0\sin (100\pi t)$ and ${x_2} = 2 \cdot 0\sin \left( {120\pi t + \dfrac{\pi }{3}} \right)$ where x is in cm and t is in second. Find the displacement of the particle at $t = 0 \cdot 0125$ .
A. -1.414
B. -1
C. 1.414
D.-2.414

Answer 1

Hint-
We can solve this question using the superposition principle. According to the superposition principle when two waves combine the net displacement will be the sum of individual displacement in both waves. So the total displacement here will be equal to the sum of ${x_1}$ and ${x_2}$ at a given time $t = 0 \cdot 0125$ .

Step by step solution:
It is given that a particle is subjected to two simple harmonic motions.
The displacement of the particle in one simple harmonic motion is given by the equation
${x_1} = 2 \cdot 0\sin (100\pi t)$
The displacement of the particle in the next simple harmonic motion is given by the equation
${x_2} = 2 \cdot 0\sin \left( {120\pi t + \dfrac{\pi }{3}} \right)$
We are asked to find the total displacement of the particle due to both these simple harmonic motions at a time $t = 0 \cdot 0125$ .
We know that according to the superposition principle when two waves combine the net displacement will be the sum of individual displacement in both waves.
So the total displacement here will be equal to the sum of ${x_1}$ and ${x_2}$.
Let the total displacement be denoted as x. then we can write ,
$x = {x_1} + {x_2}$
Now let us find the displacement due to the first simple harmonic motion at time $t = 0 \cdot 0125$.
We can find this by substituting the value of time in place of t in the equation.
${x_1} = 2 \cdot 0\sin (100\pi \times 0 \cdot 0125)$
${x_1} = 2 \cdot 0\sin (100 \times 180 \times 0 \cdot 0125)$
Since, $\pi = {180^ \circ }$
$ \Rightarrow {x_1} = 2 \cdot 0\sin \left( {225} \right)$
$ \Rightarrow {x_1} = 2 \cdot 0\sin \left( {\pi + 45} \right)$
$ \Rightarrow {x_1} = 2 \cdot 0 \times - \sin \left( {45} \right)$
$\therefore {x_1} = - \sqrt 2 $
Now let us find the displacement due to second harmonic motion.
${x_2} = 2 \cdot 0\sin \left( {120\pi \times 0 \cdot 0125 + \dfrac{\pi }{3}} \right)$
$ \Rightarrow {x_2} = 2 \cdot 0\sin \left( {120 \times 180 \times 0 \cdot 0125 + \dfrac{{180}}{3}} \right)$
$ \Rightarrow {x_2} = 2 \cdot 0\sin \left( {330} \right)$
$ \Rightarrow {x_2} = 2 \cdot 0\sin \left( {2\pi - 30} \right)$
$ \Rightarrow {x_2} = 2 \cdot 0 \times - \sin 30$
$\therefore {x_2} = - 1$
Thus, we will get the total displacement as
$x = {x_1} + {x_2} = - 1.414 + - 1$
$\therefore x = - 2.414\,cm$
This is the total displacement of the particle in time $t = 0 \cdot 0125$

So, the correct answer is option D.

Note: Remember to convert the value of $\pi $ in radian to degree before calculating the value of the trigonometric function. We know $2\pi = {360^ \circ }$ thus $\pi = {180^ \circ }$ . Also remember that $\sin \left( {{{180}^ \circ } + {x^ \circ }} \right) = - \sin {x^ \circ }$ and $\sin ({360^ \circ } - {x^ \circ }) = - \sin {x^ \circ }$