A 240-g block connected to a light spring for which the force constant is 5.50 N

Question

A 240-g block connected to a light spring for which the force constant is 5.50 N/m is free to oscillate on a frictionless, horizontal surface. The block is displaced5.20 cm from equilibrium and released from rest as in the figure.

Use the equations below. (Note that the direction is indicated by the sign in front of the equations.) x = (0.0542 m)cos(4.79t + 0.126m) --(0.259 m/s)sin(4.79t 0.126m) a =-(1.24 m/s2)cos(4.79e+ 0.126r) (e) Determine the first time (when t 0) that the position is at its maximum value (b) Determine the first time (whent that the velocity is at its maximum value.

Explanation / Answer

a)
position will be maximum when the value inside cos becomes 2*pi.
In that case cos(0) becomes 1 which is maximum.
so,
4.79*t + 0.126 pi = 2pi
4.79 t = 1.874 pi
t = 0.39*pi
= 0.39*3.14
= 1.23 s
Answer: 1.23 s

b)
velocity will be maximum when the value inside cos becomes pi.
In that case cos(pi) becomes -1 which is maximum since one - is there in front
so,
4.79*t + 0.126 pi = pi
4.79 t = 0.874 pi
t = 0.1825*pi
= 0.1825*3.14
= 0.57 s
Answer: 0.57 s

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