A 1.13-kg mass attached to a spring is displaced 7.87 cm from the equilibrium position. It is released and then oscillates with a frequency of 4.05 Hz. What is the energy of the motion when the mass passes through the equilibrium
Question
A 1.13-kg mass attached to a spring is displaced 7.87 cm from the equilibrium position. It is released and then oscillates with a frequency of 4.05 Hz. What is the energy of the motion when the mass passes through the equilibrium
Solution
The energy of the motion when the mass passes through the equilibrium can be calculated using the formula for the total energy in simple harmonic motion, which is given by:
E = 1/2 * m * ω^2 * A^2
where:
- m is the mass of the object,
- ω is the angular frequency,
- A is the amplitude of the motion.
Given:
- m = 1.13 kg,
- ω = 2π * frequency = 2π * 4.05 Hz = 25.42 rad/s,
- A = 7.87 cm = 0.0787 m.
Substituting these values into the formula gives:
E = 1/2 * 1.13 kg * (25.42 rad/s)^2 * (0.0787 m)^2 E = 0.5 * 1.13 kg * 645.2764 rad^2/s^2 * 0.00619669 m^2 E = 2.25 Joules
So, the energy of the motion when the mass passes through the equilibrium is approximately 2.25 Joules.
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