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 speed of the mass when it is 2.63 cm from the equilibrium position?
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 speed of the mass when it is 2.63 cm from the equilibrium position?
Solution
To solve this problem, we need to use the formula for the speed of an object in simple harmonic motion:
v = ω * sqrt(A^2 - x^2)
where:
- v is the speed of the object,
- ω is the angular frequency,
- A is the amplitude of the motion (the maximum displacement from the equilibrium position), and
- x is the displacement from the equilibrium position at the time we are interested in.
First, we need to convert the given values to the correct units. The displacement from the equilibrium position (A) is given as 7.87 cm, which is 0.0787 m. The displacement at the time we are interested in (x) is given as 2.63 cm, which is 0.0263 m.
Next, we need to calculate the angular frequency (ω). The frequency (f) is given as 4.05 Hz. The angular frequency is related to the frequency by the equation ω = 2πf, so ω = 2π * 4.05 = 25.46 rad/s.
Now we can substitute these values into the equation for the speed:
v = ω * sqrt(A^2 - x^2) = 25.46 * sqrt((0.0787)^2 - (0.0263)^2) = 25.46 * sqrt(0.00619689 - 0.00069169) = 25.46 * sqrt(0.0055052) = 25.46 * 0.0742 = 1.89 m/s
So, the speed of the mass when it is 2.63 cm from the equilibrium position is approximately 1.89 m/s.
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