For a wave equation y = 10 sin pi (0.01x - 2. | vedclass

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Question

(A) The given wave equation is $y = 10 \sin \pi (0.01x - 2.00t) 	ext{ cm}$.Expanding the equation,we get $y = 10 \sin (0.01\pi x - 2\pi t) 	ext{ cm}

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(A) The given wave equation is $y = 10 \sin \pi (0.01x - 2.00t) 	ext{ cm}$.Expanding the equation,we get $y = 10 \sin (0.01\pi x - 2\pi t) 	ext{ cm}$.Comparing this with the standard wave equation $y = A \sin (kx - \omega t)$,we identify the amplitude $A = 10 	ext{ cm}$ and the angular frequency $\omega = 2\pi 	ext{ rad/sec}$.The maximum particle velocity $v_{	ext{max}}$ is given by the formula $v_{	ext{max}} = A\omega$.Substituting the values,$v_{	ext{max}} = 10 	imes 2\pi = 20\pi 	ext{ cm/sec}$.Using $\pi \approx 3.14159$,we get $v_{	ext{max}} \approx 20 	imes 3.14159 = 62.83 	ext{ cm/sec}$.Rounding to the nearest integer,the maximum velocity is $63 	ext{ cm/sec}$.

For a wave equation $y = 10 \sin \pi (0.01x - 2.00t) \text{ cm}$,what is the maximum particle velocity in $\text{cm/sec}$?

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