Q189 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
A wave is represented by the equation $y=0.2\sin 4\pi\left[\dfrac{t}{0.08}+\dfrac{x}{0.8}+\dfrac{1}{12}\right]$ in SI units.
The expression represents a wave traveling in the
The expression represents a wave traveling in the
- Apositive x-direction with velocity 10 m/s and with wavelength 0.4 m.
- Correct answerBnegative x-direction with velocity 10 m/s and with wavelength 0.4 m.
- Cpositive x-direction with velocity 20 m/s and with wavelength 0.6 m.
- Dnegative x-direction with velocity 20 m/s and with wavelength 0.6 m.
Solution
The given wave equation is $y = 0.2 \sin 4\pi \left[ \dfrac{t}{0.08} + \dfrac{x}{0.8} + \dfrac{1}{12} \right]$
Comparing this with the standard wave equation $y = A \sin(\omega t + kx + \phi)$
$\omega = \dfrac{4\pi}{0.08} = 50\pi$ rad/s
$k = \dfrac{4\pi}{0.8} = 5\pi$ rad/m
Since the signs of $\omega t$ and $kx$ are the same, the wave is traveling in the negative x-direction.
The velocity of the wave is $v = \dfrac{\omega}{k} = \dfrac{50\pi}{5\pi} = 10$ m/s
The wavelength is $\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{5\pi} = 0.4$ m
Answer: negative x-direction with velocity 10 m/s and with wavelength 0.4 m.
Comparing this with the standard wave equation $y = A \sin(\omega t + kx + \phi)$
$\omega = \dfrac{4\pi}{0.08} = 50\pi$ rad/s
$k = \dfrac{4\pi}{0.8} = 5\pi$ rad/m
Since the signs of $\omega t$ and $kx$ are the same, the wave is traveling in the negative x-direction.
The velocity of the wave is $v = \dfrac{\omega}{k} = \dfrac{50\pi}{5\pi} = 10$ m/s
The wavelength is $\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{5\pi} = 0.4$ m
Answer: negative x-direction with velocity 10 m/s and with wavelength 0.4 m.
Hence, the correct answer is option: 'B'.