Question

The equation of a wave is given by $y=3sin(4x-2t)$. What is the time period and wavelength of the wave?

• A. $T = \pi, \lambda = \frac{\pi}{2}$
• B. $T = \frac{\pi}{2}, \lambda = \pi$
• C. $T = 2\pi, \lambda = \pi$
• D. $T = \pi, \lambda = 2\pi$

Answer 1

Answer: (A)

$T = \pi, \lambda = \frac{\pi}{2}$

Answer 2

The general equation of a progressive wave is given by: $y = A \sin(kx - \omega t)$

Comparing the given equation $y=3sin(4x-2t)$ with the general form:

• Amplitude $A = 3$
• Wave number $k = 4$
• Angular frequency $\omega = 2$

The relationship between angular frequency ($\omega$) and time period (T) is: $\omega = \frac{2\pi}{T}$ Therefore, $T = \frac{2\pi}{\omega}$ Substituting the value of $\omega = 2$: $T = \frac{2\pi}{2} = \pi$

The relationship between wave number (k) and wavelength ($\lambda$) is: $k = \frac{2\pi}{\lambda}$ Therefore, $\lambda = \frac{2\pi}{k}$ Substituting the value of $k = 4$: $\lambda = \frac{2\pi}{4} = \frac{\pi}{2}$

Thus, the time period of the wave is $\pi$ and the wavelength is $\frac{\pi}{2}$.