An object with mass 0.250 kg is acted on by an elastic restoring force with force constant 10.8 N/m. The object is set into oscillation with an initial potential energy of 0.150 J and an initial kinetic energy of 6.00×10−2 J.
What is the amplitude of oscillation?
What is the potential energy when the displacement is one-half the amplitude?
At what displacement are the kinetic and potential energies equal?
What is the value of the phase angle \phi if the initial velocity is positive and the initial displacement is negative?
(1/2) k x^2 + (1/2)mV^2 = 0.21 J
= total energy = constant
Set V = 0 to solve for the amplitude, since V = 0 then
When displacement is half the amplitude, potential energy is 1/4 of the total, 0.1575 J
When KE = PE,
(1/2) kx^2 = 0.015 J. Solve for x at that time
The phase angle depends upon what you call zero phase. Yopu haven't said
Why did the object go to the party alone?
Because it wanted to oscillate freely without any attachments!
Jokes aside, let's solve your questions:
1. To find the amplitude of oscillation, we can use the equation for total mechanical energy:
E = (1/2)kA^2
where E is the total mechanical energy, k is the force constant, and A is the amplitude of oscillation.
From the given information, we know that the initial potential energy (E_p) is 0.150 J and the initial kinetic energy (E_k) is 6.00x10^-2 J.
The total mechanical energy can be found by summing the initial potential and kinetic energies:
E = E_p + E_k = 0.150 J + 6.00x10^-2 J = 0.210 J
Substituting this into the equation above, we can solve for A:
0.210 J = (1/2)(10.8 N/m)A^2
A^2 = (2)(0.210 J) / (10.8 N/m)
A^2 ≈ 0.038888889 m^2
A ≈ √(0.038888889) m
A ≈ 0.197 m
Therefore, the amplitude of oscillation is approximately 0.197 m.
2. To find the potential energy when the displacement is one-half the amplitude, we can use the equation for potential energy in simple harmonic motion:
E_p = (1/2)kx^2
where E_p is the potential energy, k is the force constant, and x is the displacement from the equilibrium position.
Given that the displacement is one-half the amplitude (A/2), we have:
E_p = (1/2)(10.8 N/m)(0.197 m/2)^2
E_p = (1/2)(10.8 N/m)(0.197 m)^2 / 4
E_p ≈ 0.0534805 J
Therefore, when the displacement is one-half the amplitude, the potential energy is approximately 0.0534805 J.
3. To find the displacement where the kinetic and potential energies are equal, we can equate the equations for kinetic energy (E_k) and potential energy (E_p):
(1/2)kx^2 = (1/2)mv^2
where x is the displacement, m is the mass of the object, and v is the velocity.
Given that the mass of the object is 0.250 kg and the initial kinetic energy is 6.00x10^-2 J, we can write:
(1/2)(10.8 N/m)x^2 = (1/2)(0.250 kg)(v^2)
Dividing both sides by (1/2):
(10.8 N/m)x^2 = (0.250 kg)(v^2)
Squaring both sides and substituting v = ωA (where ω is the angular frequency):
(10.8 N/m)x^2 = (0.250 kg)((2πf)^2)(A^2)
Simplifying:
x^2 = (0.250 kg)((2πf)^2)(A^2) / (10.8 N/m)
Since we are looking for the displacement where the kinetic and potential energies are equal, we can set E_k = E_p and substitute the given values:
(1/2)kx^2 = (1/2)mv^2
(1/2)kx^2 = (1/2)(0.250 kg)(6.00x10^-2 J)
(1/2)(10.8 N/m)x^2 = (1/2)(0.250 kg)(6.00x10^-2 J)
(10.8 N/m)x^2 = (0.250 kg)(6.00x10^-2 J)
x^2 = (0.250 kg)(6.00x10^-2 J) / (10.8 N/m)
x^2 ≈ 0.006944444 m^2
x ≈ √(0.006944444) m
x ≈ 0.083333333 m
Therefore, at a displacement of approximately 0.083333333 m, the kinetic and potential energies are equal.
4. To find the phase angle (φ) given a positive initial velocity and negative initial displacement, we can use the equation for displacement in simple harmonic motion:
x(t) = A cos(ωt + φ)
where x(t) is the displacement at time t, A is the amplitude, ω is the angular frequency, and φ is the phase angle.
Given that the initial velocity (v) is positive and initial displacement (x) is negative, we can conclude that the object is moving in the positive direction at the start, but is displaced in the negative direction.
Since cosine is positive in the first and fourth quadrants, this tells us that the phase angle should be in the second quadrant.
Therefore, the value of the phase angle φ is approximately π radians (180 degrees).
I hope you found this oscillating response amusing!
To find the amplitude of oscillation, we can use the equation for the total mechanical energy of the system:
E = (1/2)kA^2
where E is the initial total mechanical energy, k is the force constant, and A is the amplitude of oscillation.
Given that the initial potential energy is 0.150 J, the initial kinetic energy is 6.00×10^−2 J, and the force constant is 10.8 N/m, we can substitute these values into the equation to solve for A:
E = (1/2)kA^2
0.150 J + 6.00×10^−2 J = (1/2)(10.8 N/m)(A^2)
0.21 J = 5.4 N/m * A^2
A^2 = 0.21 J / 5.4 N/m
A^2 ≈ 0.038889
Taking the square root of both sides gives:
A ≈ √0.038889
A ≈ 0.197 m
Therefore, the amplitude of oscillation is approximately 0.197 m.
To find the potential energy when the displacement is one-half the amplitude, we can use the equation for potential energy:
PE = (1/2)kx^2
where PE is the potential energy, k is the force constant, and x is the displacement.
Since x is one-half the amplitude, x = 0.5A = 0.5(0.197 m) = 0.0985 m. We can substitute these values into the equation to find PE:
PE = (1/2)kx^2
PE = (1/2)(10.8 N/m)(0.0985 m)^2
PE ≈ 0.053 J
Therefore, the potential energy when the displacement is one-half the amplitude is approximately 0.053 J.
To find the displacement at which the kinetic and potential energies are equal, we can set the equations for kinetic and potential energies equal to each other:
(1/2)mv^2 = (1/2)kx^2
where m is the mass, v is the velocity, k is the force constant, and x is the displacement.
Given that the mass is 0.250 kg and the force constant is 10.8 N/m, we can set up the equation:
(1/2)(0.250 kg)v^2 = (1/2)(10.8 N/m)x^2
0.125 kg v^2 = 5.4 N/m x^2
v^2 = 5.4 N/m x^2 / 0.125 kg
v^2 = 43.2 N/m x^2 / 0.25 kg
v^2 = 20.16 N/m x^2
v = sqrt(20.16 N/m) * x
v = 4.495 x
Since the kinetic and potential energies are equal, the v^2 terms cancel out:
1 = sqrt(20.16 N/m) * x
To simplify further, we can square both sides and solve for x:
1 = (20.16 N/m) x^2
x^2 = 1 / (20.16 N/m)
x^2 ≈ 0.049603
x ≈ sqrt(0.049603)
x ≈ 0.222 m
Therefore, the displacement at which the kinetic and potential energies are equal is approximately 0.222 m.
To find the value of the phase angle φ, we can use the definition of the phase angle:
tan(φ) = (initial displacement) / (initial velocity)
Given that the initial velocity is positive and the initial displacement is negative, we can substitute these values into the equation:
tan(φ) = (negative initial displacement) / (positive initial velocity)
Since both the initial displacement and initial velocity are positive in magnitude, the phase angle φ will be in the second quadrant, meaning it will be negative.
Therefore, the value of the phase angle φ if the initial velocity is positive and the initial displacement is negative is a negative value.
To find the amplitude of oscillation, you can use the formula for the total mechanical energy of a mass-spring system:
E = (1/2)kA^2,
where E is the total mechanical energy, k is the force constant, and A is the amplitude.
Given that the initial potential energy (PE) is 0.150 J and the initial kinetic energy (KE) is 6.00×10^(-2) J, you can calculate the total mechanical energy:
E = PE + KE,
E = 0.150 J + 6.00×10^(-2) J,
E = 0.210 J.
Now, set this equal to the formula for the total mechanical energy:
0.210 J = (1/2)(10.8 N/m)A^2.
Solving for the amplitude (A), we get:
A^2 = (2)(0.210 J) / (10.8 N/m),
A^2 = 0.039 J / N,
A = √(0.039 J / N),
A ≈ 0.198 m.
Therefore, the amplitude of oscillation is approximately 0.198 meters.
Next, to find the potential energy when the displacement is one-half the amplitude, you can use the equation for potential energy:
PE = (1/2)kX^2,
where PE is the potential energy, k is the force constant, and X is the displacement.
Since the displacement is one-half the amplitude (X = A/2), we can calculate the potential energy:
PE = (1/2)(10.8 N/m)(A/2)^2,
PE = (1/8)(10.8 N/m)A^2,
PE = (1/8)(10.8 N/m)(0.198 m)^2,
PE ≈ 0.027 J.
Therefore, the potential energy when the displacement is one-half the amplitude is approximately 0.027 Joules.
Now, to find the displacement at which the kinetic and potential energies are equal, you can set the equations for kinetic and potential energy equal to each other:
(1/2)kX^2 = (1/2)(mv^2),
where X is the displacement, k is the force constant, m is the mass, and v is the velocity.
Given that the initial kinetic energy is 6.00×10^(-2) J, you can express the kinetic energy in terms of mass and velocity:
(1/2)(mv^2) = 6.00×10^(-2) J,
mv^2 = 0.12 J,
v^2 = 0.12 J / m.
Now, substitute the expression for velocity into the equation:
(1/2)kX^2 = 0.12 J / m,
kX^2 = 0.24 J / m,
X^2 = 0.24 J / (m*k),
X = √(0.24 J / (m*k)).
To find the value of X, you would need the mass of the object. Once you have the mass, plug in the values into the equation above to get the displacement at which the kinetic and potential energies are equal.
Finally, the phase angle (phi) relates to the initial velocity (v) and the initial displacement (X) through the equation:
tan(phi) = -vX / (√(k/m)).
If the initial velocity is positive and the initial displacement is negative, you can substitute these values in the equation:
tan(phi) = -(positive initial velocity) * (negative initial displacement) / (√(k/m)).
Note that if the initial velocity is positive and the initial displacement is negative, the product -(positive initial velocity) * (negative initial displacement) would be positive since two negatives would make a positive. Then you can calculate the value of phi by calculating the arctan of the ratio:
phi = arctan(-(positive initial velocity) * (negative initial displacement) / (√(k/m))).
To find the value of phi, you would need the values for the initial velocity, the initial displacement, and the mass. Once you have those values, plug them into the equation above to get the value of the phase angle.