A transverse wave is represented by y =2 sin (ω t - kx ) cm. The value of wavelength (in cm ) for which the wave velocity becomes equal to the maximum particle velocity, will be ;

Question

A transverse wave is represented by y =2 sin (ω t - kx ) cm. The value of wavelength (in cm ) for which the wave velocity becomes equal to the maximum particle velocity, will be ; (A) 4 π (B) 2 π (C) π (D) 2

Tardigrade Admin · Accepted Answer

y=2 sin (ω t-k x) Maximum particle velocity = text A ω Wave velocity =(ω/ k ) (ω/ k )= A ω k =(1/ A )=(2 π/λ) λ=2 π A =4 π cm

Q. A transverse wave is represented by $y = 2 sin (ω t - k x) c m$ . The value of wavelength (in $c m$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be ;

$4 π$

$2 π$

$π$

2

$y = 2 sin (ω t - k x)$
Maximum particle velocity $= A ω$
Wave velocity $= \frac{ω}{k}$
$\frac{ω}{k} = A ω$
$k = \frac{1}{A} = \frac{2 π}{λ}$
$λ = 2 π A$
$= 4 π c m$

Q. A transverse wave is represented by y=2sin(ωt−kx)cm. The value of wavelength (in cm ) for which the wave velocity becomes equal to the maximum particle velocity, will be ;

4π

2π

π

2

y=2sin(ωt−kx) Maximum particle velocity = A ω Wave velocity =kω​ kω​=Aω k=A1​=λ2π​ λ=2πA =4πcm

Q. A transverse wave is represented by $y = 2 sin (ω t - k x) c m$ . The value of wavelength (in $c m$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be ;

$4 π$

$2 π$

$π$

$y = 2 sin (ω t - k x)$
Maximum particle velocity $= A ω$
Wave velocity $= \frac{ω}{k}$
$\frac{ω}{k} = A ω$
$k = \frac{1}{A} = \frac{2 π}{λ}$
$λ = 2 π A$
$= 4 π c m$