A transverse wave is described by the equation y = A 2π (nt - (x)/(10)). The ratio of maximum particle velocity to wave velocity is

Correct answer: A. (π A)/(5) The given wave equation is y = A 2π ( nt - (x)/(10)) Comparing this with the standard wave equation y = A ( ω t - kx) , we get: Angular frequency, ω =2π n Wave number, k = (2π)/(10) = (π)/(5) Maximum particle velocity is given by v_ p,max = A ω =A(2π n) Wave velocity is given by v_w = (ω)/(k) = (2π n)/(π/5) =10n The ratio of maximum particle velocity to wave velocity is: v_ p,max v_w = (A(2π n))/(10n) = (π A)/(5) Answer: (π A)/(5)

MHT CET 202621 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Morning Shift
Subject: Physics
Chapter: Waves and Sound

MHT CET 2026 · 21 April · Morning Shift · Physics · Waves and Sound

A transverse wave is described by the equation y = A 2π (nt -...

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

Options

A $\dfrac{\pi A}{5}$
B $\dfrac{2\pi A}{5}$
C $\dfrac{3\pi A}{5}$
D $\dfrac{4\pi A}{5}$

Correct answer

A. $\dfrac{\pi A}{5}$

Step-by-step solution

The given wave equation is

y = A \sin 2\pi \left(

nt -

\dfrac{x}{10}\right)<br><br>Comparing this with the standard wave equation

y = A \sin(

\omega

t - kx)

, we get:<br><br>Angular frequency, \omega =2\pi

<br><br>Wave number,

k =

\dfrac{2\pi}{10} = \dfrac{\pi}{5}<br><br>Maximum particle velocity is given by

v_{p,max} = A

\omega =A(2\pi

<br><br>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}<br><br>Answer: \dfrac{\pi A}{5}

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