A block with a mass of 0.306 kg is attached to a spring of spring constant 444 N/m. It is sitting at equilibrium. You then pull the block down 4.50 cm from equilibrium and let go. What is the amplitude of the oscillation?
To find the amplitude of the oscillation, we need to understand the relationship between the spring constant, mass, and amplitude in simple harmonic motion.
In simple harmonic motion, the displacement (x) of an object attached to a spring is given by the equation:
x = A * sin(ωt)
Where:
- x is the displacement from equilibrium,
- A is the amplitude (maximum displacement),
- ω is the angular frequency, and
- t is the time.
The angular frequency (ω) can be calculated using the formula:
ω = √(k/m)
Where:
- k is the spring constant, and
- m is the mass of the object.
Given that the mass is 0.306 kg and the spring constant is 444 N/m, we can calculate the angular frequency (ω).
ω = √(444 N/m / 0.306 kg)
≈ √(1450.98)
≈ 38.08 rad/s
Now, we can find the displacement at any given time by substituting the values of ω and t into the equation. However, to find the amplitude, we need to consider the initial conditions.
Since the block is pulled down 4.50 cm from equilibrium and then released, we can assume that at t = 0, the displacement x is equal to 4.50 cm or 0.045 m.
0.045 m = A * sin(ω * 0)
Since sin(0) = 0, we can conclude that A * 0 = 0.045 m.
Therefore, the amplitude of the oscillation is 0.045 m or 4.50 cm.