What is the relation between particle velocity (v), the wave velocity (c) and the slope (s) of the wave?

Question

What is the relation between particle velocity (v), the wave velocity (c) and the slope (s) of the wave? (A)  v=-cs  (B)  c=-vs  (C)  v2=c2s  (D)  c2=v2s

Tardigrade Admin · Accepted Answer

Let the wave equation y=A sin (ω t-kx) then v=(dy/dt)Aω cos (ω t-kx) and s=(dy/dx)Ak cos (ω t-kx) =(-kv/ω )=-(v/c) v=-cs

Q. What is the relation between particle velocity (v), the wave velocity (c) and the slope (s) of the wave?

$v = - cs$

$c = - v s$

$v^{2} = c^{2} s$

$c^{2} = v^{2} s$

Let the wave equation
$y = A sin (ω t - k x)$
then $v = \frac{d y}{d t} A ω cos (ω t - k x)$
and $s = \frac{d y}{d x} A k cos (ω t - k x)$
$= \frac{- k v}{ω} = - \frac{v}{c}$
$v = - cs$

Q. What is the relation between particle velocity (v), the wave velocity (c) and the slope (s) of the wave?

v=−cs

c=−vs

v2=c2s

c2=v2s

Let the wave equation y=Asin(ωt−kx) then v=dtdy​Aωcos(ωt−kx) and s=dxdy​Akcos(ωt−kx) =ω−kv​=−cv​ v=−cs

$v = - cs$

$c = - v s$

$v^{2} = c^{2} s$

$c^{2} = v^{2} s$

Let the wave equation
$y = A sin (ω t - k x)$
then $v = \frac{d y}{d t} A ω cos (ω t - k x)$
and $s = \frac{d y}{d x} A k cos (ω t - k x)$
$= \frac{- k v}{ω} = - \frac{v}{c}$
$v = - cs$