Energy of Simple Harmonic Oscillators - AP Physics C: Mechanics
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Calculate the kinetic energy when $x = A/2$ for a simple harmonic oscillator.
Calculate the kinetic energy when $x = A/2$ for a simple harmonic oscillator.
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$K = \frac{3}{4} \times E$. At $x = A/2$, kinetic energy is $E - U = E - \frac{E}{4} = \frac{3E}{4}$.
$K = \frac{3}{4} \times E$. At $x = A/2$, kinetic energy is $E - U = E - \frac{E}{4} = \frac{3E}{4}$.
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State the phase difference between velocity and displacement in simple harmonic motion.
State the phase difference between velocity and displacement in simple harmonic motion.
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$\frac{\pi}{2}$ radians. Velocity leads displacement by 90 degrees in SHM.
$\frac{\pi}{2}$ radians. Velocity leads displacement by 90 degrees in SHM.
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If total energy is $5 \text{ J}$ and $k = 100 \text{ N/m}$, find the amplitude.
If total energy is $5 \text{ J}$ and $k = 100 \text{ N/m}$, find the amplitude.
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$A = 0.316 \text{ m}$. From $E = \frac{1}{2}kA^2$, solving for $A = \sqrt{\frac{2E}{k}} = 0.316$ m.
$A = 0.316 \text{ m}$. From $E = \frac{1}{2}kA^2$, solving for $A = \sqrt{\frac{2E}{k}} = 0.316$ m.
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State the formula for the kinetic energy of a simple harmonic oscillator.
State the formula for the kinetic energy of a simple harmonic oscillator.
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$K = \frac{1}{2} m v^2$. Standard kinetic energy formula applies to oscillating mass.
$K = \frac{1}{2} m v^2$. Standard kinetic energy formula applies to oscillating mass.
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Find the total energy if $k = 50 \text{ N/m}$ and $A = 0.2 \text{ m}$.
Find the total energy if $k = 50 \text{ N/m}$ and $A = 0.2 \text{ m}$.
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$E = 1 \text{ J}$. Using $E = \frac{1}{2}kA^2 = \frac{1}{2}(50)(0.2)^2 = 1$ J.
$E = 1 \text{ J}$. Using $E = \frac{1}{2}kA^2 = \frac{1}{2}(50)(0.2)^2 = 1$ J.
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What remains constant in an ideal simple harmonic oscillator without friction?
What remains constant in an ideal simple harmonic oscillator without friction?
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Total mechanical energy. Energy conservation applies in ideal oscillator systems.
Total mechanical energy. Energy conservation applies in ideal oscillator systems.
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Find the potential energy when $x = A$ in a simple harmonic oscillator.
Find the potential energy when $x = A$ in a simple harmonic oscillator.
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$U = \frac{1}{2} k A^2$. At amplitude, displacement equals maximum, so potential energy is maximum.
$U = \frac{1}{2} k A^2$. At amplitude, displacement equals maximum, so potential energy is maximum.
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Calculate the total mechanical energy for $m = 1 \text{ kg}$, $v_{max} = 2 \text{ m/s}$.
Calculate the total mechanical energy for $m = 1 \text{ kg}$, $v_{max} = 2 \text{ m/s}$.
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$E = 2 \text{ J}$. Total energy equals maximum kinetic energy: $\frac{1}{2}mv_{max}^2 = 2$ J.
$E = 2 \text{ J}$. Total energy equals maximum kinetic energy: $\frac{1}{2}mv_{max}^2 = 2$ J.
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State the energy conservation principle in simple harmonic motion.
State the energy conservation principle in simple harmonic motion.
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Total energy is conserved. Kinetic plus potential energy remains constant.
Total energy is conserved. Kinetic plus potential energy remains constant.
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Find the potential energy when $x = A$ in a simple harmonic oscillator.
Find the potential energy when $x = A$ in a simple harmonic oscillator.
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$U = \frac{1}{2} k A^2$. At amplitude, displacement equals maximum, so potential energy is maximum.
$U = \frac{1}{2} k A^2$. At amplitude, displacement equals maximum, so potential energy is maximum.
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What is the role of damping in a real-world oscillator system?
What is the role of damping in a real-world oscillator system?
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Reduces total energy over time. Damping causes energy loss through friction or resistance.
Reduces total energy over time. Damping causes energy loss through friction or resistance.
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Determine the potential energy at the midpoint of displacement for an oscillator.
Determine the potential energy at the midpoint of displacement for an oscillator.
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$U = \frac{1}{4} \times E$. At $x = A/2$, potential energy is $\frac{1}{2}k(\frac{A}{2})^2 = \frac{E}{4}$.
$U = \frac{1}{4} \times E$. At $x = A/2$, potential energy is $\frac{1}{2}k(\frac{A}{2})^2 = \frac{E}{4}$.
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What is the kinetic energy at maximum displacement?
What is the kinetic energy at maximum displacement?
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Zero. At maximum displacement, all energy is potential.
Zero. At maximum displacement, all energy is potential.
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Identify the expression for maximum kinetic energy in a simple harmonic oscillator.
Identify the expression for maximum kinetic energy in a simple harmonic oscillator.
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$K_{max} = \frac{1}{2} k A^2$. Maximum kinetic energy equals total mechanical energy.
$K_{max} = \frac{1}{2} k A^2$. Maximum kinetic energy equals total mechanical energy.
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Identify the energy form at the equilibrium position in SHM.
Identify the energy form at the equilibrium position in SHM.
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Maximum kinetic energy. At equilibrium, all energy is kinetic since potential energy is zero.
Maximum kinetic energy. At equilibrium, all energy is kinetic since potential energy is zero.
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What is the potential energy if the displacement is zero?
What is the potential energy if the displacement is zero?
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Zero. At equilibrium position, spring is neither compressed nor extended.
Zero. At equilibrium position, spring is neither compressed nor extended.
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Calculate the total mechanical energy for $m = 1 \text{ kg}$, $v_{max} = 2 \text{ m/s}$.
Calculate the total mechanical energy for $m = 1 \text{ kg}$, $v_{max} = 2 \text{ m/s}$.
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$E = 2 \text{ J}$. Total energy equals maximum kinetic energy: $\frac{1}{2}mv_{max}^2 = 2$ J.
$E = 2 \text{ J}$. Total energy equals maximum kinetic energy: $\frac{1}{2}mv_{max}^2 = 2$ J.
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Calculate the frequency if $\text{ω} = 6.28 \text{ rad/s}$.
Calculate the frequency if $\text{ω} = 6.28 \text{ rad/s}$.
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$f = 1 \text{ Hz}$. Using $f = \frac{\omega}{2\pi} = \frac{6.28}{2\pi} = 1$ Hz.
$f = 1 \text{ Hz}$. Using $f = \frac{\omega}{2\pi} = \frac{6.28}{2\pi} = 1$ Hz.
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State what happens to energy distribution at maximum velocity.
State what happens to energy distribution at maximum velocity.
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All energy is kinetic. At maximum velocity, displacement is zero so only kinetic energy exists.
All energy is kinetic. At maximum velocity, displacement is zero so only kinetic energy exists.
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State the expression for maximum potential energy in a simple harmonic oscillator.
State the expression for maximum potential energy in a simple harmonic oscillator.
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$U_{max} = \frac{1}{2} k A^2$. Maximum potential energy equals total mechanical energy.
$U_{max} = \frac{1}{2} k A^2$. Maximum potential energy equals total mechanical energy.
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Find the potential energy if $k = 200 \text{ N/m}$ and $x = 0.1 \text{ m}$.
Find the potential energy if $k = 200 \text{ N/m}$ and $x = 0.1 \text{ m}$.
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$U = 1 \text{ J}$. Using $U = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.1)^2 = 1$ J.
$U = 1 \text{ J}$. Using $U = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.1)^2 = 1$ J.
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Find the total mechanical energy if $k = 100 \text{ N/m}$ and $A = 0.5 \text{ m}$.
Find the total mechanical energy if $k = 100 \text{ N/m}$ and $A = 0.5 \text{ m}$.
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$E = 12.5 \text{ J}$. Using $E = \frac{1}{2}kA^2 = \frac{1}{2}(100)(0.5)^2 = 12.5$ J.
$E = 12.5 \text{ J}$. Using $E = \frac{1}{2}kA^2 = \frac{1}{2}(100)(0.5)^2 = 12.5$ J.
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Calculate the kinetic energy if $m = 2 \text{ kg}$ and $v = 3 \text{ m/s}$.
Calculate the kinetic energy if $m = 2 \text{ kg}$ and $v = 3 \text{ m/s}$.
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$K = 9 \text{ J}$. Using $K = \frac{1}{2}mv^2 = \frac{1}{2}(2)(3)^2 = 9$ J.
$K = 9 \text{ J}$. Using $K = \frac{1}{2}mv^2 = \frac{1}{2}(2)(3)^2 = 9$ J.
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Determine the frequency if $k = 250 \text{ N/m}$ and $m = 2 \text{ kg}$. Round to 2 decimals.
Determine the frequency if $k = 250 \text{ N/m}$ and $m = 2 \text{ kg}$. Round to 2 decimals.
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$f \thickapprox 1.78 \text{ Hz}$. Using $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{250}{2}} \approx 1.78$ Hz.
$f \thickapprox 1.78 \text{ Hz}$. Using $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{250}{2}} \approx 1.78$ Hz.
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What is the relationship between kinetic and potential energy at maximum displacement?
What is the relationship between kinetic and potential energy at maximum displacement?
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Kinetic energy is zero. All energy is potential at maximum displacement.
Kinetic energy is zero. All energy is potential at maximum displacement.
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Which energy is zero when the oscillator passes through the equilibrium position?
Which energy is zero when the oscillator passes through the equilibrium position?
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Potential energy. All energy is kinetic at equilibrium position.
Potential energy. All energy is kinetic at equilibrium position.
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State the phase difference between velocity and displacement in simple harmonic motion.
State the phase difference between velocity and displacement in simple harmonic motion.
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$\frac{\text{π}}{2}$ radians. Velocity leads displacement by 90 degrees in SHM.
$\frac{\text{π}}{2}$ radians. Velocity leads displacement by 90 degrees in SHM.
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What is the phase difference between acceleration and displacement in simple harmonic motion?
What is the phase difference between acceleration and displacement in simple harmonic motion?
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$\text{π}$ radians. Acceleration is opposite to displacement in SHM.
$\text{π}$ radians. Acceleration is opposite to displacement in SHM.
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What is the phase relationship between potential energy and displacement in SHM?
What is the phase relationship between potential energy and displacement in SHM?
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In phase. Both potential energy and displacement squared vary together.
In phase. Both potential energy and displacement squared vary together.
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Calculate the maximum velocity if $A = 0.3 \text{ m}$ and $T = 2 \text{ s}$. Round to 2 decimals.
Calculate the maximum velocity if $A = 0.3 \text{ m}$ and $T = 2 \text{ s}$. Round to 2 decimals.
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$v_{max} \thickapprox 0.94 \text{ m/s}$. Using $v_{max} = \frac{2\pi A}{T} = \frac{2\pi(0.3)}{2} \approx 0.94$ m/s.
$v_{max} \thickapprox 0.94 \text{ m/s}$. Using $v_{max} = \frac{2\pi A}{T} = \frac{2\pi(0.3)}{2} \approx 0.94$ m/s.
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