An object with mass 5.0 kg is attached to a spring with spring stiffness constant 192 N/m and is executing simple harmonic motion. When the object is 0.35 m from its equilibrium position, it is moving with a speed of 0.02 m/s. Calculate the amplitude of the motion.
To solve this problem, we can use the equation for the velocity of an object undergoing simple harmonic motion at a distance x from its equilibrium position:
v = ω√(A^2 - x^2)
where v is the velocity, ω is the angular frequency, A is the amplitude of the motion, and x is the distance from the equilibrium position.
We are given v = 0.02 m/s and x = 0.35 m. We need to find the amplitude A.
First, let's find the angular frequency using the equation:
ω = √(k/m)
where k is the spring constant and m is the mass.
Given k = 192 N/m and m = 5.0 kg, we can calculate ω:
ω = √(192 N/m / 5.0 kg)
= √(38.4 N/kg)
≈ 6.21 rad/s
Now we can substitute the values of v, ω, and x into the equation for velocity:
0.02 m/s = 6.21 rad/s * √(A^2 - (0.35 m)^2)
Square both sides to isolate the A term:
(0.02 m/s)^2 = (6.21 rad/s)^2 * (A^2 - (0.35 m)^2)
0.0004 m^2/s^2 = 38.5841 rad^2/s^2 * (A^2 - 0.1225 m^2)
Divide both sides by 38.5841 rad^2/s^2:
0.0004 m^2/s^2 / 38.5841 rad^2/s^2 = A^2 - 0.1225 m^2
A^2 = 0.0004 m^2/s^2 / 38.5841 rad^2/s^2 + 0.1225 m^2
A^2 ≈ 0.00001036 m^2 + 0.1225 m^2
A^2 ≈ 0.12251036 m^2
Taking the square root of both sides, we find:
A ≈ √(0.12251036 m^2)
≈ 0.35 m (rounded to two decimal places)
Therefore, the amplitude of the motion is approximately 0.35 m.
To calculate the amplitude of the motion, we need to use the equation for the total mechanical energy of an object undergoing simple harmonic motion:
E = (1/2)kx² + (1/2)mv²
Where:
E is the total mechanical energy
k is the spring stiffness 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, the total mechanical energy is the sum of the potential energy stored in the spring and the kinetic energy of the object.
Since we are given the displacement x and the velocity v at that displacement, we can solve for the amplitude A. At the extreme positions of simple harmonic motion, the object is momentarily at rest, so the velocity is zero.
At the given displacement:
x = 0.35 m
v = 0.02 m/s
Knowing the spring stiffness constant:
k = 192 N/m
We can calculate the amplitude using the formula for total mechanical energy:
E = (1/2)kx² + (1/2)mv²
Since the velocity v is zero at the extreme positions, we can ignore the kinetic energy term:
E = (1/2)kx²
Now we can plug in the values to solve for the total mechanical energy:
E = (1/2)(192 N/m)(0.35 m)²
E = (1/2)(192 N/m)(0.1225 m²)
E = (1/2)(23.52 J*m)
E = 11.76 J*m
The total mechanical energy of the system is 11.76 J*m.
Now, we can use the equation for total mechanical energy to solve for the amplitude A:
E = (1/2)kA²
Plugging in the values:
11.76 J*m = (1/2)(192 N/m)A²
Rearranging the equation to solve for A:
A² = (2 * 11.76 J*m) / (192 N/m)
A² = 23.52 J*m / 192 N/m
A² = 0.1225 m²
Taking the square root of both sides:
A = √(0.1225 m²)
A = 0.35 m
Therefore, the amplitude of the motion is 0.35 m.