A transverse stationary wave passes through a string with the equation y =10 sin π(0.02 x -2.00 t ) where x is in meters and t in seconds. The maximum velocity of the particles in wave motion is

Question

A transverse stationary wave passes through a string with the equation y =10 sin π(0.02 x -2.00 t ) where x is in meters and t in seconds. The maximum velocity of the particles in wave motion is (A) 63 (B) 78 (C) 100 (D) 121

Tardigrade Admin · Accepted Answer

y=10 sin π(0.02 x-2.00 t) ( partial y/ partial t)=-20 π cos π(0.02 x-2.00 t) (( partial y/ partial t) ÷) max =20 π=63 units.

Q. A transverse stationary wave passes through a string with the equation $y = 10 sin π (0.02 x - 2.00 t)$ where $x$ is in meters and $t$ in seconds. The maximum velocity of the particles in wave motion is

63

78

100

121

$y = 10 sin π (0.02 x - 2.00 t)$
$\frac{\partial y}{\partial t} = - 20 π cos π (0.02 x - 2.00 t)$
$(\frac{\partial y}{\partial t} \div)_{m a x} = 20 π = 63$ units.

Q. A transverse stationary wave passes through a string with the equation y=10sinπ(0.02x−2.00t) where x is in meters and t in seconds. The maximum velocity of the particles in wave motion is

63

78

100

121

y=10sinπ(0.02x−2.00t) ∂t∂y​=−20πcosπ(0.02x−2.00t) (∂t∂y​÷)max​=20π=63 units.

Q. A transverse stationary wave passes through a string with the equation $y = 10 sin π (0.02 x - 2.00 t)$ where $x$ is in meters and $t$ in seconds. The maximum velocity of the particles in wave motion is

$y = 10 sin π (0.02 x - 2.00 t)$
$\frac{\partial y}{\partial t} = - 20 π cos π (0.02 x - 2.00 t)$
$(\frac{\partial y}{\partial t} \div)_{m a x} = 20 π = 63$ units.