A massless spring with k = 206 N/m is hung from a support and a 1.31 kg mass is
ID: 1287773 • Letter: A
Question
A massless spring with k = 206 N/m is hung from a support and a 1.31 kg mass is attached to the bottom of the spring. The position of the mass when it is in equilibrium is y = 0. If the mass is displaced below this point, we take its displacement to be positive. At t = 0, the mass has the initial conditions y(0) = 0.1 m and [dy/dt]t=0= -1.7 m/s.
A) What is the value of A?
B) What is the value of ??
C) What is the value of ??
A massless spring with k = 206 N/m is hung from a support and a 1.31 kg mass is attached to the bottom of the spring. The position of the mass when it is in equilibrium is y = 0. If the mass is displaced below this point, we take its displacement to be positive. At t = 0, the mass has the initial conditions y(0) = 0.1 m and [dy/dt]t=0= -1.7 m/s. Its position at later times is: y(t)=A cos (wt+delta) A) What is the value of A? B) What is the value of ?? C) What is the value of ??Explanation / Answer
y(t) = Acos(wt + &)
y(0) = Acos(&) =0.1
y'(t) = Aw sin(wt + &)
y'(0) = Aw sin(&) = -1.7
w = sqrt(k/m) = 12.54 ........Ans
A sin& = -1.7/12.54 = -0.136
A sin& / Acos& = tan& = -1.36
& = - 53.59 dgrees ..........Ans
sin& = -0.805
A = -0.136 / sin& = 0.169 m ..........Ans
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.