MasterinqPh,sics Ch15 HW - CoogleChrome ü Serure https://sesan.masteringphysics.
ID: 1655039 • Letter: M
Question
MasterinqPh,sics Ch15 HW - CoogleChrome ü Serure https://sesan.masteringphysics.com/myctitemView?assignmentproblemlD=81830462 Ch 15 HW Problem 15.25 Rescurces T previous 10 of 15 next Problem 15.25 Submit My Answers Glve Up Correct The let-hand end of a long horizontal stretched cord oscillates transversely in SHM with frequency 300 Iz and amplitude 2.6 " The cord is under a tension af 190 N and has a linear density 0.09 kg/m . At t = 0, the end of the cord has an upward cisplacement of 1.5 cn and is falling. Consider the point = 1.35 m on the oord/Figure 1) Part C What is its velocity at 1 = 3.0 s ? Express your answer to two significant figures and include the appropriate units. u= 1 Value Units Figure 1 Submit My Answers Give Up Part D what is its acceleration at t = 3.0 s ? Wave velocity Express your answer to two significant figures and include the appropriate units. Han Value Units Submit My Answers Give Up Provide Feedback Continue Type here to search 8:42 PM 9/16 2017Explanation / Answer
v=sqrt(T/) = sqrt(190/0.09) = 45.95 m/s
=v/f = 45.95/300 = 0.153 m
Let us first find the wave equation,
D(x,t) = Asin(kx-t+)
A=0.026m, k=2/ = (2)/0.153 = 41.07 , = 2 f = 2*3.14*300 = 1884 s^-1
At t= 0, 0.015 = 0.026sin
= 0.615 rad
Plugging in above equation,
D(x,t) = Asin(kx-t+)
D(x,t) = 0.026sin(41.07x-1884t+0.615)
a)
D(x,t) = 0.026sin(41.07x-1884t+0.615)
v(x,t) = d/dt[D(x,t)]
v(x,t) =d/dt[0.026sin(41.07x-1884t+0.615)]
v(x,t) = -48.98cos(41.07x-1884t+0.615)
Put x=1.35m and t=3.0s
v(x,t) = -48.98cos(41.07*1.35-1884*3.0+0.615)
v(x,t)=35 m/s
b)
a(x,t) =d/dt[-48.98cos(41.07*x-1884*t+0.615)
a(x,t)=-92278.3sin(41.07x-1884t+0.615)
Put x=1.35m and t=3.0s
a(x,t)=-92278.3sin(41.07*1.35-1884*3.0+0.615)
a(x,t) = -64000 m/s^2