Question

Identify, which of above is a wave motion. find velocity of wave.

• A. y = 4 cos(200$\pi$t+10$\pi$x)
• B. y = 6 tan[20$\pi$t-$\pi$x]
• C. y = $\frac{20}{4+(6x-18t)^2+(4x-2t)^2}$
• D. y = 8 $e$

Answer 1

Answer: (A), (B)

Option (a) is a wave motion with velocity $\overrightarrow{v}$ = -20$\hat{i}$ m/s.

Answer 2

A wave motion is described by an equation of the form $y = f(ax \pm bt)$. The velocity of such a wave is given by $v = |b/a|$.

Option (a): $y = 4 \cos(200\pi t + 10\pi x)$. This is of the form $y = A \cos(\omega t + kx)$. Here, $\omega = 200\pi$ and $k = 10\pi$. The wave velocity magnitude is $v = \omega/k = \frac{200\pi}{10\pi} = 20$ m/s. The term $(kx + \omega t)$ indicates propagation in the negative x-direction. Thus, the velocity vector is $\overrightarrow{v_a} = -20\hat{i}$ m/s. This matches the given $\overrightarrow{v}$.

Option (b): $y = 6 \tan[20\pi t - \pi x]$. This is of the form $y = A \tan(\omega t - kx)$. Here, $\omega = 20\pi$ and $k = \pi$. The wave velocity magnitude is $v = \omega/k = \frac{20\pi}{\pi} = 20$ m/s. The term $(\omega t - kx)$ indicates propagation in the positive x-direction. Thus, the velocity vector is $\overrightarrow{v_b} = +20\hat{i}$ m/s. This does not match the given $\overrightarrow{v}$.

Option (c): $y = \frac{20}{4+(6x-18t)^2+(4x-2t)^2}$. This equation is not of the form $y = f(ax \pm bt)$ and does not represent a simple wave motion.

Option (d): $y = 8 e$. This is a constant and does not represent a wave motion.

Therefore, option (a) represents the wave motion with the velocity matching the given vector $\overrightarrow{v} = -20\hat{i}$ m/s.