14.34.Solve:The object’s position as a function of time is ( ) ( )0cos.xtAtωφ=+Letting 0 mx=at 0 s,t=gives 10020cosAφφπ=⇒=±Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between −and 0 radians. Thus, 102=−and ()( ) ( )( )1122cossin0.10 m sinxtAtxtA ttπω⇒==where we have used 0.10 m
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