Question

The equation $y=A{\sin }^2(kx-\omega t)$ represents a wave with

• A. amplitude $A$ frequency $\omega /2\mathrm{\pi }$
• B. amplitude $A/2$ frequency $\omega /\mathrm{\pi }$
• C. amplitude $2A$ frequency $\omega /4\mathrm{\pi }$
• D. It does not represent a wave motion

Answer 1

The correct option is B

amplitude $A/2$ frequency $\omega /\mathrm{\pi }$

GIven:

The equation of the wave is $y=A{\sin }^2(kx-\omega t)$

Step 1: convert the equation into the standard wave equation form

Formula : ${\sin }^2\theta =\frac{1}{2}(1-2\cos 2\theta )$

⇒ $y=\frac{A}{2}(1-\cos (2kx-2\omega t\left)\right)$

⇒ The main wave equation will be $y=\frac{A}{2}\left(\cos \right(2kx-2\omega t\left)\right)$

Step 2: Compare the main equation

The wave equation is $y=A{\sin }^2(kx-\omega t)$

Frequency $v=\frac{\omega }{2\pi }$ , the frequency of our equation is $v\text{'}=\frac{2\omega }{2\pi }$

Amplitude is $A$, here the amplitude is$\frac{A}{2}$

Thus, The frequency and amplitude of the given wave are$\frac{A}{2}$ and $\frac{\omega }{\pi }$ respectively.

Hence, the correct answer is option B