A massless spring with spring constant 17.1 N/m hangs vertically. A body of mass 0.210 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. What is the amplitude of the resulting motion?
To find the amplitude of the resulting motion, we can use the equation for the period of a mass-spring system:
T = 2π√(m/k)
Where:
T = period of the motion
m = mass of the body (0.210 kg)
k = spring constant (17.1 N/m)
First, we need to find the period of the motion. The period is the time it takes for the system to complete one full oscillation. In this case, since the body is hanging vertically, it will undergo simple harmonic motion.
T = 2π√(m/k)
T = 2π√(0.210 kg / 17.1 N/m)
Now, let's plug in the values and solve for T:
T = 2π√(0.210 kg / 17.1 N/m)
T ≈ 2π√(0.0123 s²/kg)
Using the equation T = 2π√(0.0123 s²/kg), we can solve for T:
T ≈ 2π * √(0.0123 s²/kg)
T ≈ 2π * 0.111 s/kg
T ≈ 0.698 s
Now that we have the period of the motion (T), we can find the amplitude using the formula:
A = (2π / T) * √((mg) / k)
Where:
A = amplitude of the motion
m = mass of the body (0.210 kg)
g = acceleration due to gravity (approximately 9.8 m/s²)
k = spring constant (17.1 N/m)
Let's calculate the amplitude:
A = (2π / T) * √((mg) / k)
A ≈ (2π / 0.698 s) * √((0.210 kg * 9.8 m/s²) / 17.1 N/m)
Plugging in the values:
A ≈ (2π / 0.698 s) * √(2.0586 kg*m/s² / 17.1 N/m)
A ≈ (2π / 0.698 s) * √(0.120479 m)
A ≈ (2π / 0.698 s) * 0.347 m
A ≈ 15π / 34 ≈ 1.449∼1.45 m
Therefore, the amplitude of the resulting motion is approximately 1.45 meters.