Question

A transverse wave is described by the equation y = A sin 2π [ t/T - x/λ]. For which wavelength of a wave, maximum particle velocity is two times the wave velocity?
(a) λ = πA/4
(b) λ = πA/2
(c) λ = πA
(d) λ = 2πA

Answer 1

Answer:

option (b) right hai this is done in notebook

Answer 2

Answer:

(c) λ = πA

Explanation:

As per the question,

Given:

A transverse wave equation :

$y=Asin2\pi[\frac{t}{T}-\frac{x}{\lambda}]$

From this equation, we have

Amplitude = A

T = 1/f

Where f = frequency

The maximum velocity of particle is given by

V(max) = Amplitude × 2πf

∴ V(max) = 2πfA

Velocity of the wave is given by:

V(wave) = λ × f

Now, from the question

The maximum particle velocity is two times the wave velocity, that is

V(max) = 2 × V(wave)

⇒ 2πfA = 2 × λ × f

⇒ λ = πA

Hence, the required wavelength = πA