Q162 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
A transverse wave is described by the equation $y = A \sin 2\pi\left(nt - \dfrac{x}{10}\right)$. The ratio of maximum particle velocity to wave velocity is
- Correct answerA$\dfrac{\pi A}{5}$
- B$\dfrac{2\pi A}{5}$
- C$\dfrac{3\pi A}{5}$
- D$\dfrac{4\pi A}{5}$
Solution
The given wave equation is $y = A \sin 2\pi\left(nt - \dfrac{x}{10}\right)$
Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$, we get:
Angular frequency, $\omega = 2\pi n$
Wave number, $k = \dfrac{2\pi}{10} = \dfrac{\pi}{5}$
Maximum particle velocity is given by $v_{p,max} = A\omega = A(2\pi n)$
Wave velocity is given by $v_w = \dfrac{\omega}{k} = \dfrac{2\pi n}{\pi/5} = 10n$
The ratio of maximum particle velocity to wave velocity is:
$\dfrac{v_{p,max}}{v_w} = \dfrac{A(2\pi n)}{10n} = \dfrac{\pi A}{5}$
Answer: $\dfrac{\pi A}{5}$
Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$, we get:
Angular frequency, $\omega = 2\pi n$
Wave number, $k = \dfrac{2\pi}{10} = \dfrac{\pi}{5}$
Maximum particle velocity is given by $v_{p,max} = A\omega = A(2\pi n)$
Wave velocity is given by $v_w = \dfrac{\omega}{k} = \dfrac{2\pi n}{\pi/5} = 10n$
The ratio of maximum particle velocity to wave velocity is:
$\dfrac{v_{p,max}}{v_w} = \dfrac{A(2\pi n)}{10n} = \dfrac{\pi A}{5}$
Answer: $\dfrac{\pi A}{5}$
Hence, the correct answer is option: 'A'.