The problem uses the ideal spring force, in one dimension. F = -kx, where x is the deviation from equilibrium, and k is the spring constant. Calculate the particle's maximum speed (vmax), with these numbers: k = 2000 N/m = 2000 kg/s2 m = 5 kg (the mass of the particle) x0 = 0.2 m (the initial position) v0 = 5 m/s (the initial velocity) The all-too-common erroneous solution: v max = A ω = x 0 k / m = 0.2 m ⋅ 20 / s = 4 m/s . Your problem: You may not understand this problem fully, and may not be able to solve it. Nevertheless, spot the egregious wrongness of the answer.
The problem uses the ideal spring force, in one dimension. F = -kx, where x is the deviation from equilibrium, and k is the spring constant. Calculate the particle's maximum speed (vmax), with these numbers: k = 2000 N/m = 2000 kg/s2 m = 5 kg (the mass of the particle) x0 = 0.2 m (the initial position) v0 = 5 m/s (the initial velocity) The all-too-common erroneous solution: v max = A ω = x 0 k / m = 0.2 m ⋅ 20 / s = 4 m/s . Your problem: You may not understand this problem fully, and may not be able to solve it. Nevertheless, spot the egregious wrongness of the answer.
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The problem uses the ideal spring force, in one dimension. F = -kx, where x is the deviation from equilibrium, and k is the spring constant. Calculate the particle's maximum speed (vmax), with these numbers:
- k = 2000 N/m = 2000 kg/s2
- m = 5 kg (the mass of the particle)
- x0 = 0.2 m (the initial position)
- v0 = 5 m/s (the initial velocity)
The all-too-common erroneous solution: v max = A ω = x 0 k / m = 0.2 m ⋅ 20 / s = 4 m/s .
Your problem: You may not understand this problem fully, and may not be able to solve it. Nevertheless, spot the egregious wrongness of the answer.
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