Question

The amplitude and frequency of a wave represented by the equation:
\[Y=2A{{\cos }^{2}}(k\gamma -\omega t)\] are –
\[\begin{align}
  & \text{A) 2A, }\dfrac{\omega }{2\pi } \\
 & \text{B) A, }\dfrac{\omega }{\pi } \\
 & \text{C) }\sqrt{A},\dfrac{\omega }{\pi } \\
 & \text{D) }\sqrt{A},\dfrac{\omega }{2\pi } \\
\end{align}\]

Answer 1

Hint: We need to know the standard equations involved in a Simple Harmonic equation to compare the given equation and analyse the different parameters involved in the two equations to get the correct value of amplitude and frequency.

Complete step-by-step answer:
We know that the equation of motion of a particle in a Simple Harmonic Motion is a sinusoidal function depending on the angular frequency of motion, the amplitude, the phase constant and other factors.
The equation of motion of a simple harmonic motion is given in standard form as –
\[y=a\cos (k\gamma -\omega t+\phi )\]
Where, y is the displacement of the particle from the mean position at any time ‘t’,
a is the maximum amplitude of oscillation,
k is the propagation constant,
\[\gamma \] is the direction of wave propagation,
\[\omega \] is the angular frequency of the SHM,
\[\phi \] is the phase difference, if any.
Now, we can compare the equation of motion given to us. It is given as –
\[Y=2A{{\cos }^{2}}(k\gamma -\omega t)\]
We can see that the wave equation is represented as the square of the cosine function. We can use the trigonometric relations for converting the square of the cosine to the cosine function. The following relation can be used.
\[\begin{align}
  & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2} \\
\end{align}\]
Now, let us convert the given function into the terms of cosine using this relation as –
\[\begin{align}
  & Y=2A{{\cos }^{2}}(k\gamma -\omega t) \\
 & \Rightarrow Y=\dfrac{2A}{2}(1+\cos 2(k\gamma -\omega t)) \\
 & \therefore Y=A(1+\cos (2k\gamma -2\omega t)) \\
\end{align}\]
Now, when we compare the standard equation with the equation given to us, we can see that –
\[\begin{align}
  & \therefore a=A \\
 & \text{and,} \\
 & \therefore f=\dfrac{2\omega }{2\pi }=\dfrac{\omega }{\pi } \\
\end{align}\]
We get the amplitude and the frequency as above.
The correct answer is option B.

So, the correct answer is “Option B”.

Note: We have to be careful while dealing with the equations of motion of simple harmonic motion. Because the equivalent equations can be really convincing as of the standard form which can result in highly ambiguous outcomes of amplitude and frequency.