Question
How does kinetic energy relate to potential energy when an object's displacement is half the minimum value in simple harmonic motion?
Answers
Answered by
GPT-4o mini
In simple harmonic motion (SHM), the relationship between kinetic energy (KE) and potential energy (PE) can be analyzed using the concepts of displacement, amplitude, and conservation of energy.
In SHM, the total mechanical energy \( E \) of the system is conserved and is given by the sum of kinetic and potential energy:
\[
E = KE + PE
\]
The potential energy in SHM is typically expressed as:
\[
PE = \frac{1}{2} k x^2
\]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. The kinetic energy is given by:
\[
KE = \frac{1}{2} m v^2
\]
where \( m \) is the mass and \( v \) is the velocity.
The maximum displacement from equilibrium in SHM is known as the amplitude \( A \). When an object is at a displacement \( x \), the total energy in the system can also be expressed as:
\[
E = \frac{1}{2} k A^2
\]
Given this, we can find the energies at specific displacements. If the displacement is half the amplitude, \( x = \frac{A}{2} \):
1. Calculate the potential energy at this displacement:
\[
PE = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{2} k \left(\frac{A^2}{4}\right) = \frac{1}{8} k A^2
\]
2. Now, since the total mechanical energy \( E \) is \( \frac{1}{2} k A^2 \):
\[
E = KE + PE \implies \frac{1}{2} k A^2 = KE + \frac{1}{8} k A^2
\]
3. Rearranging gives the kinetic energy at that displacement:
\[
KE = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \left(\frac{4}{8} - \frac{1}{8}\right) k A^2 = \frac{3}{8} k A^2
\]
Therefore, when the object's displacement is half the minimum value (amplitude), the potential energy is \( \frac{1}{8} k A^2 \) and the kinetic energy is \( \frac{3}{8} k A^2 \). This demonstrates how kinetic energy and potential energy can be distributed in SHM based on displacement.
The total energy in the system remains constant and is shared between kinetic and potential forms as the object oscillates.
In SHM, the total mechanical energy \( E \) of the system is conserved and is given by the sum of kinetic and potential energy:
\[
E = KE + PE
\]
The potential energy in SHM is typically expressed as:
\[
PE = \frac{1}{2} k x^2
\]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. The kinetic energy is given by:
\[
KE = \frac{1}{2} m v^2
\]
where \( m \) is the mass and \( v \) is the velocity.
The maximum displacement from equilibrium in SHM is known as the amplitude \( A \). When an object is at a displacement \( x \), the total energy in the system can also be expressed as:
\[
E = \frac{1}{2} k A^2
\]
Given this, we can find the energies at specific displacements. If the displacement is half the amplitude, \( x = \frac{A}{2} \):
1. Calculate the potential energy at this displacement:
\[
PE = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{2} k \left(\frac{A^2}{4}\right) = \frac{1}{8} k A^2
\]
2. Now, since the total mechanical energy \( E \) is \( \frac{1}{2} k A^2 \):
\[
E = KE + PE \implies \frac{1}{2} k A^2 = KE + \frac{1}{8} k A^2
\]
3. Rearranging gives the kinetic energy at that displacement:
\[
KE = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \left(\frac{4}{8} - \frac{1}{8}\right) k A^2 = \frac{3}{8} k A^2
\]
Therefore, when the object's displacement is half the minimum value (amplitude), the potential energy is \( \frac{1}{8} k A^2 \) and the kinetic energy is \( \frac{3}{8} k A^2 \). This demonstrates how kinetic energy and potential energy can be distributed in SHM based on displacement.
The total energy in the system remains constant and is shared between kinetic and potential forms as the object oscillates.