Question

The displacement of a particle executing simple harmonic motion is given by y\[ = {A_0} + A\sin \omega t + B\cos \omega t\]. Then the amplitude of its oscillation is given by:
A. \[{A_0} + \sqrt {{A^2} + {B^2}} \]
B. \[\sqrt {{A^2} + {B^2}} \]
C. \[\sqrt {{A_0}^2 + {{(A + B)}^2}} \]
D. \[A + B\]

Answer 1

Hint: To solve this question, we have to know what is simple harmonic motion. From the name we can say this is a motion where the restoring force on the moving particle or object is directly proportional to the magnitude of the displacement which acts to the object’s equilibrium position which is the mean position. We can say, the maximum displacement on one side is exactly equal to the displacement of the other side. A force is responsible for this simple harmonic motion.

Complete step by step solution:
we know, according to the above question, y\[ = {A_0} + A\sin \omega t + B\cos \omega t\]
From this equation we can say, here two waves are superimposing on one another.so, here we can say the phase difference between two waves is \[\pi /2\].
Equation of simple harmonic motion is, \[y' = y - {A_0} = A\sin \omega t + B\cos \omega t\]
So, we can write now the resultant amplitude is,
R is equal to \[\sqrt {{A^2} + {B^2} + 2AB\cos \Delta } \phi \]
\[\Delta \phi \] is equal to the phase difference.
Putting the values we can write,
\[\sqrt {{A^2} + {B^2} + 2AB\cos \pi /2} = R\]
$R = \sqrt {{A^2} + {B^2} + 2AB.0} \\
\therefore R = \sqrt {{A^2} + {B^2}} $

So, the right option will be option B.

Note: We can get confused between simple harmonic motion and damped vibration. In simple harmonic motion we know that the time interval of every complete vibration is the same or constant. We can write, \[F = - Kx\]. Here, F is the force which is responsible for this motion and x is the displacement caused by the motion. K is the constant. This is called Hooke’s law.