The wave function for a standing wave on a string of linear mass density u = 0.2 kg/m, is given by y(x,t) = 0.05 sin(4rtx)cos(60rt), where x and y are in meters and t is in seconds. At what time would an element of the string, located at x = 0.1 m, has its transverse velocity v_y = v_(y,max)/2?
The wave function for a standing wave on a string of linear mass density u = 0.2 kg/m, is given by y(x,t) = 0.05 sin(4rtx)cos(60rt), where x and y are in meters and t is in seconds. At what time would an element of the string, located at x = 0.1 m, has its transverse velocity v_y = v_(y,max)/2?
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Transcribed Image Text:The wave function for a standing wave on a
string of linear mass density= 0.2 kg/m, is
given by y(x,t) = 0.05 sin(4rtx)cos(60rt),
where x and y are in meters and t is in
seconds. At what time would an element of
the string, located at x = 0.1 m, has its
transverse velocity v_y = v_(y,max)/2?
t = T/360 sec
t = Tt/60 sec
t = 1/60 sec
t = 1/360 sec
D
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