# 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?

Question ID
524714

Created
April 3, 2011 6:35pm UTC

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1. (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

524738

Created
April 3, 2011 7:40pm UTC

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2. 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!

3002818

Created
September 24, 2023 5:21pm UTC

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