A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)
The speed of the object at the instant when its displacement is 0.040 m can be calculated using the conservation of energy equation.
Speed = sqrt(2*(0.5*k*(0.080 m)^2 - 0.5*k*(0.040 m)^2))/0.20 kg
Speed = sqrt(2*(0.8 - 0.2))/0.20 kg
Speed = sqrt(0.6)/0.20 kg
Speed = 1.2 m/s
To find the speed of the object at a given displacement, we can use the conservation of energy principle. The total mechanical energy of the system remains constant throughout the motion.
The total mechanical energy of a mass-spring system is given by the equation:
E = 0.5 * k * A^2
where E is the total mechanical energy, k is the spring constant, and A is the amplitude of motion.
In this case, the total mechanical energy E is equal to the potential energy at the equilibrium position, when the displacement is zero.
So, setting the potential energy equal to the total mechanical energy:
0.5 * k * A^2 = 0.5 * k * x^2
where x is the displacement at the instant we want to find the speed (0.040 m in this case).
Simplifying the equation, we have:
0.5 * k * A^2 = 0.5 * k * x^2
0.5 * 10 N/m * (0.080 m)^2 = 0.5 * 10 N/m * (0.040 m)^2
1 J = 0.4 J
Since the total mechanical energy remains constant, the kinetic energy will be equal to the total mechanical energy at other positions. Therefore, we can equate the kinetic energy to the total mechanical energy to find the speed.
0.5 * m * v^2 = E
0.5 * 0.20 kg * v^2 = 0.4 J
Multiplying both sides by 2, we have:
0.20 kg * v^2 = 0.8 J
Dividing both sides by 0.20 kg, we get:
v^2 = 4 J/kg
Taking the square root of both sides, we find:
v = √(4 J/kg)
v = 2 m/s
Therefore, the speed of the object at the instant when its displacement is 0.040 m is 2 m/s.
To find the speed of the object at a specific displacement during simple harmonic motion, we can use the conservation of energy principle.
The total mechanical energy of the object in simple harmonic motion is constant and can be expressed as the sum of its potential energy and its kinetic energy:
E = 1/2 k x^2 + 1/2 m v^2
Where:
- E is the total mechanical energy.
- k is the spring constant.
- x is the displacement from the equilibrium position.
- m is the mass of the object.
- v is the velocity of the object.
In this case, we need to find the speed at a specific displacement of 0.040 m.
First, we need to determine the total mechanical energy of the system. At the maximum displacement (amplitude), the object is at its maximum potential energy and has zero kinetic energy. Therefore, the total mechanical energy is:
E = 1/2 k A^2 + 0
E = 1/2 * 10 N/m * (0.080 m)^2
E = 0.032 J
Next, we can use this value of mechanical energy to find the speed at a displacement of 0.040 m.
E = 1/2 k x^2 + 1/2 m v^2
Substituting the known values:
0.032 J = 1/2 * 10 N/m * (0.040 m)^2 + 1/2 * 0.20 kg * v^2
0.032 J = 0.008 J + 1/2 * 0.20 kg * v^2
Simplifying the equation:
0.032 J - 0.008 J = 1/2 * 0.20 kg * v^2
0.024 J = 0.10 kg * v^2
0.024 J / 0.10 kg = v^2
0.24 m^2/s^2 = v^2
Taking the square root of both sides:
v = √(0.24 m^2/s^2)
v ≈ 0.49 m/s
Therefore, at the instant when the displacement is 0.040 m, the speed of the object is approximately 0.49 m/s.