A 200g block is connected to a light spring having a constant of 5.0 N/m is free

Question

A 200g block is connected to a light spring having a constant of 5.0 N/m is free to oscillate on a frictionless horizontal surface. The block is displaced 5.0cm from the equilibrium position and released from rest.

a) find period of motion

b)determine the maximum speed of the block

c)what is the maximum acceleration of the block

d) express the postion, velocity, and acceleration as a function of time (in Si units)

e)If the block were released from same initial position, x=5.0cm but with an initial velocity of v=-0.1 m/s what is the new period of motion, maximum speed and amplitude

I really would like the anwser for [part e] especially thanks!

Explanation / Answer

m = 0. = 0.2 kg

k = 5 N/m

x = 0.05 m

w = sqrt(k/m) = 5 rad/s

a) T = 2*pi/w = 6.28/5 = 1.256 s

b) Vmax = A*w = 0.25 m/s

c) a_max = A*w^2 = 1.25 m/s^2

d) x = A*cos(w*t)

v = dx/dt = -A*w*sin(wt)

a = dV/dt = -A*w^2*cos(w*t)

e) Time periode remains same

let A is the amplitude

0.5*k*A^2 = 0.5*m*v^2 + 0.5*k*x^2

5*A^2 = 0.2*0.1^2 + 5*0.05^2

A = 0.0539 m = 5.39 cm

Vmax = A*w = 0.2695 m/s

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