Damped Oscillations. Mass Hanging from a spring

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Damping is negligible for a 0.121 kg mass on a 6.55 N/m spring, and a driving force of 1.45 N is applied. To find the frequency that results in an amplitude of 0.465 m, the equation for a damped harmonic oscillator can be simplified by ignoring the damping term. By rearranging the equation, the frequency can be calculated using the formula ω = sqrt((1.45N) / (0.121kg * 0.465m)) - (6.55N/m) / (2 * 0.121kg). This results in two possible frequency solutions: ω = 4.57 rad/s or ω = -5.72 rad/s.
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Damping is negligible for a 0.121 kg mass hanging from a light 6.55 N/m spring. The system is driven by a force oscillating with an amplitude of 1.45 N. At what frequency will the force make the mass vibrate with an amplitude of 0.465 m? There are two possible solutions, enter one of them.


Now, I realize that damping is negligible so the part of the equation (whichever one it is) that includes the damping force will equal 0.

However I'm unsure where to begin this problem or how to relate the driving force to the 0.465m amplitude.

Any help on where to get started?
 
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The equation for a damped harmonic oscillator is given by: F(t) = -k * x(t) - c * v(t). Where F(t) is the driving force, k is the spring constant, x(t) is the displacement and v(t) is the velocity. Using the given parameters, you can rearrange the equation to solve for the frequency of the oscillations: ω = sqrt((1.45N) / (0.121kg * 0.465m)) - (6.55N/m) / (2 * 0.121kg) This yields two possible solutions, ω = 4.57 rad/s or ω = -5.72 rad/s.
 
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