Posted by **srikala** on Sunday, May 27, 2012 at 1:50pm.

the equation of a transverse wave travel ling on a rope is given by y=10sin3.14(0.01x-2.00t)where y and x are in cm and t is in seconds.the maximum transverse speed of the particle in the rope is about

options:

1. 62.8 cm/s

2. 75 cm/s

- physics -
**Damon**, Sunday, May 27, 2012 at 4:09pm
dy/dt = -10*2*3.14 cos****

max of cos is 1

so 20*3.14 = 62.8

- physics -
**Elena**, Sunday, May 27, 2012 at 4:24pm
y=10sin 3.14(0.01x-2.00t)=

= 10sin (π•x/100 - 2•π•t).

General form is

y= Asin (2π•x/λ - ω•t).

v =dy/dt =- A •ω cos (2•π•x/λ - ω•t).

v(max) = A •ω,where A =10 cm, ω = 2•π (rad/s),

v(max) = 20• π =62.8 cm/s

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