A 4.05 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 121 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 8.65 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
To find the amplitude of the resulting simple harmonic motion, we need to first determine the maximum displacement of the block from its equilibrium position.
The maximum displacement is equal to the amplitude, which can be calculated using Hooke's law and the energy conservation principle.
First, let's find the potential energy stored in the spring when it is stretched. The potential energy stored in a spring can be calculated using the formula:
Potential energy (PE) = (1/2) * k * x^2
Where:
k = spring constant = 121 N/m (given)
x = displacement from equilibrium position (unknown)
Next, let's find the kinetic energy of the block when it is shoved with an initial speed:
Kinetic energy (KE) = (1/2) * m * v^2
Where:
m = mass of the block = 4.05 x 10^-2 kg (given)
v = initial speed = 8.65 m/s (given)
Now, according to the principle of energy conservation, the total mechanical energy (E) of the system (spring and block) is conserved. Therefore, the sum of potential energy and kinetic energy is constant:
E = PE + KE
Since the spring is initially unstrained, the total mechanical energy is equal to the kinetic energy:
E = KE
Now, let's substitute the values into the equation:
E = (1/2) * m * v^2
E = (1/2) * (4.05 x 10^-2 kg) * (8.65 m/s)^2
Solve the equation to find the value of E.
Once we have the value of E, we can find the displacement (x) using the potential energy equation:
PE = (1/2) * k * x^2
Rearrange the equation:
x = √(2 * PE / k)
Substitute the values of PE and k into the equation:
x = √(2 * E / k)
Solve the equation to find the displacement (x). This displacement will give us the amplitude of the simple harmonic motion.
Note: The amplitude in simple harmonic motion is defined as half of the peak-to-peak displacement.
To find the amplitude of the resulting simple harmonic motion, we can use the formula for the potential energy of a spring:
𝑃𝑒 = 1/2 𝑘𝑥²
where 𝑃𝑒 is the potential energy, 𝑘 is the spring constant, and 𝑥 is the displacement from the equilibrium position.
In this case, since the spring is initially unstrained, the displacement 𝑥 will be equal to the amplitude (A) of the simple harmonic motion.
We can find the potential energy using the initial conditions given:
𝑃𝑒 = 1/2 𝑘𝑥²
𝑃𝑒 = 1/2 * 121 N/m * (4.05 x 10^(-2) kg * (8.65 m/s)²)
Now we can solve for the amplitude (A):
𝑥² = 2𝑃𝑒/𝑘
𝐴 = √(2𝑃𝑒/𝑘)
Plugging in the values, we have:
𝐴 = √(2 * (1/2 * 121 N/m * (4.05 x 10^(-2) kg * (8.65 m/s)²)) / 121 N/m)
Simplifying the equation:
𝐴 = √(1 * (4.05 x 10^(-2) kg * (8.65 m/s)²))
𝐴 = √(4.05 x 10^(-2) kg * (8.65 m/s)²)
𝐴 ≈ √(3.4763 x 10^(-2) kg * m²/s²)
𝐴 ≈ 0.186 m
Therefore, the amplitude of the resulting simple harmonic motion is approximately 0.186 meters.