A block of mass m = 5 kg connected to a massless spring undergoes simple harmonic motion (SHM) on a horizontal frictionless surface. When released at t = 0 the block starts at xo = +0.4 m with speed vo = 0. The block then reaches its maximum speed vmax at x = 0 when t = π/4 s.
Since we are not given K in this problem, I don't know where to even start. This is not a HW problem. This is a practice problem for my final on Wednesday. Thank you!!
a) (5 points) Find the amplitude A of this SHM.
b) (5 points) Find the period T of this SHM.
c) (6 points) Find the spring constant k.
d) (6 points) Use conservation of energy to find vmax.
e) (7 points) Write down the general expression for x(t), the position at any time t for this SHM, and find the position x at time t = 2π s
3
f) (6 points) Find the shortest time that it takes for the block to travel from x = +0.8A to x = -0.6A.
There is no diagram included with this problem.
Since we are not given K in this problem, I don't know where to even start. This is not a HW problem. This is a practice problem for my final on Wednesday. Thank you!!
a) (5 points) Find the amplitude A of this SHM.
b) (5 points) Find the period T of this SHM.
c) (6 points) Find the spring constant k.
d) (6 points) Use conservation of energy to find vmax.
e) (7 points) Write down the general expression for x(t), the position at any time t for this SHM, and find the position x at time t = 2π s
3
f) (6 points) Find the shortest time that it takes for the block to travel from x = +0.8A to x = -0.6A.
There is no diagram included with this problem.
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I will do a few of these...
a) the amplitude is the maximum displacment, so is 0.4m
b) the time from max amplitude to equilibrium position is 1/4 the period; the max speed occurs at equilibrium (where all PE is converted to KE); so the period is 4 x pi/4 or pi secs
c) the period of an oscillator = 2 pi Sqrt[m/k]
you know the period = pi and m = 5kg, so solve for k
d) knowing k, find vmax from conservation of energy:
initial PE = KE at equilibrium
1/2 k x^2 = 1/2 m vmax^2
you know all the values except vmax, so solve for vmax
e) x(t) = 0.4 cos(2t)
when t = 2 pi, x = 0.4 m
a) the amplitude is the maximum displacment, so is 0.4m
b) the time from max amplitude to equilibrium position is 1/4 the period; the max speed occurs at equilibrium (where all PE is converted to KE); so the period is 4 x pi/4 or pi secs
c) the period of an oscillator = 2 pi Sqrt[m/k]
you know the period = pi and m = 5kg, so solve for k
d) knowing k, find vmax from conservation of energy:
initial PE = KE at equilibrium
1/2 k x^2 = 1/2 m vmax^2
you know all the values except vmax, so solve for vmax
e) x(t) = 0.4 cos(2t)
when t = 2 pi, x = 0.4 m