1) A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harmonic motion with an amplitude of 10 cm. What is the speed of the 1.0 kg mass at the equilibrium position?
To find the speed of the 1.0 kg mass at the equilibrium position, we can use the equation for the total mechanical energy in simple harmonic motion:
E = (1/2) * k * x^2
Where:
- E is the total mechanical energy
- k is the force constant of the spring
- x is the displacement from the equilibrium position
In this case, the displacement from the equilibrium position is 0 since the mass is at the equilibrium position. Therefore, the equation simplifies to:
E = (1/2) * k * 0^2 = 0
Since the total mechanical energy is zero at the equilibrium position, the kinetic energy is also zero. Therefore, the speed of the 1.0 kg mass at the equilibrium position is zero.
To find the speed of the 1.0 kg mass at the equilibrium position, we need to understand a few concepts related to simple harmonic motion and springs.
First, we need to know the relationship between the force exerted by a spring and the displacement from the equilibrium position. Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, it can be expressed as:
F = -kx
Where:
F is the force exerted by the spring,
k is the force constant (also known as the spring constant),
x is the displacement from the equilibrium position.
Since the mass is set in simple harmonic motion, we know that the force exerted by the spring follows a sinusoidal pattern. At the equilibrium position (where x = 0), there is no force acting on the mass.
Now, let's find the equilibrium position. The amplitude of the motion is given as 10 cm. The amplitude is the maximum displacement from the equilibrium position. In this case, the equilibrium position is at x = 0.
Next, let's find the force constant of the spring. According to the problem statement, the force constant is 400 N/m.
Given that the mass (m) is 1.0 kg and the force constant (k) is 400 N/m, we can calculate the speed of the mass at the equilibrium position.
The total mechanical energy of the system (E) is the sum of the potential energy (U) and the kinetic energy (K):
E = U + K
At the equilibrium position, the potential energy is at its maximum and the kinetic energy is zero. Therefore, the total mechanical energy is equal to the potential energy:
E = U
The potential energy of a mass-spring system is given by the formula:
U = (1/2)kx^2
Since the mass is at the equilibrium position, x = 0, and hence the potential energy U will be zero.
Therefore, the total mechanical energy E is zero.
The kinetic energy (K) of an object can be calculated using the formula:
K = (1/2)mv^2
Where:
m is the mass of the object,
v is its velocity.
Since the total mechanical energy E is zero, we can equate it to the sum of the potential energy U and the kinetic energy K:
E = U + K
0 = 0 + (1/2)mv^2
Simplifying the equation, we get:
0 = (1/2)mv^2
Since the mass (m) is non-zero (1.0 kg), the velocity (v) must be zero in order to satisfy the equation.
Therefore, the speed of the 1.0 kg mass at the equilibrium position is zero.