dy/dt = -10*2*3.14 cos****
max of cos is 1
so 20*3.14 = 62.8
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
2 answers
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
= 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
1 answer