Asked by Cp
A wave on a string is described by:
y=(3.6cm)sin(2.7x−3.1t +0.97)
where x is in meters, and t is in seconds, phase constant is in radians
Consider a particle at x= 0.50m. After
t = 0, how much time will pass before that particle first reaches its maximum transverse speed?
I tried to find the time with using the maximum transverse speed but I don't know how to find the distance
y=(3.6cm)sin(2.7x−3.1t +0.97)
where x is in meters, and t is in seconds, phase constant is in radians
Consider a particle at x= 0.50m. After
t = 0, how much time will pass before that particle first reaches its maximum transverse speed?
I tried to find the time with using the maximum transverse speed but I don't know how to find the distance
Answers
Answered by
Steve
transverse speed is dy/dt, so x is constant.
dy/dt = (3.6cm)(-3.1)cos(2.7x−3.1t +0.97)
at x = 0.5, that is
dy/dt = -11.16 cos(2.32-3.1t)
To find maximum speed, you need
y" = 11.16(-3.1) sin(2.32-3.1t) = 0
clearly that is at 2.32-3.1t=0 or pi. t = 0.75 or 1.76
we want a max, not a min, so t=1.76
You can see from the graph that this is so.
http://www.wolframalpha.com/input/?i=3.6+sin(2.7*0.5%E2%88%923.1t+%2B0.97)
dy/dt = (3.6cm)(-3.1)cos(2.7x−3.1t +0.97)
at x = 0.5, that is
dy/dt = -11.16 cos(2.32-3.1t)
To find maximum speed, you need
y" = 11.16(-3.1) sin(2.32-3.1t) = 0
clearly that is at 2.32-3.1t=0 or pi. t = 0.75 or 1.76
we want a max, not a min, so t=1.76
You can see from the graph that this is so.
http://www.wolframalpha.com/input/?i=3.6+sin(2.7*0.5%E2%88%923.1t+%2B0.97)
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