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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/physics/edexcel/as-energy-in-oscillations/flashcards","label":"Energy in oscillations","subjectCode":"as_level_physics","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"5f8bcdb2-01d9-467c-a5df-7c2c4f2a73ef","title":"Energy in oscillations Flashcards","topicId":"e171d40a-7075-5308-b145-1ef1c8a53e88","topicSlug":"as-energy-in-oscillations","cards":[{"id":"378b5243-d0e5-4954-a388-d070b9da548c","cardType":"basic","points":1,"front":"What is the resultant force on an object at its equilibrium position?","back":"Zero","mcqOptions":[{"text":"Zero","correct":true,"explanation":"At the equilibrium position, displacement is zero. Since the restoring force in simple harmonic motion is proportional to displacement, the resultant force is also zero."},{"text":"Maximum","correct":false,"explanation":"Maximum resultant force occurs at the extreme positions, where displacement is greatest. At equilibrium, displacement and therefore the restoring force are zero."},{"text":"Equal to the weight","correct":false,"explanation":"This confuses an individual force with the resultant force. At equilibrium, forces such as weight and tension balance, so the resultant force is zero."},{"text":"Equal to the restoring force","correct":false,"explanation":"The resultant force is the restoring force, but at equilibrium the displacement is zero, making that restoring force and hence the resultant force zero."}]},{"id":"f10fb5f6-3de9-408b-9eaa-5780810d2a98","cardType":"basic","points":2,"front":"Describe the energy stores at the extreme positions ($x = \\pm A$) of an oscillation.","back":"

Kinetic energy is zero
Potential energy is maximum

","mcqOptions":[{"text":"

Kinetic energy is zero
Potential energy is maximum

","correct":true,"explanation":"At either extreme position, the oscillator stops momentarily before reversing direction, so its kinetic energy is zero and its potential energy is maximum."},{"text":"

Kinetic energy is maximum
Potential energy is zero

","correct":false,"explanation":"This reverses the energy distribution at an extreme position. The oscillator is instantaneously stationary there, so kinetic energy is zero and potential energy is maximum."},{"text":"

Kinetic energy is maximum
Potential energy is maximum

","correct":false,"explanation":"Both stores cannot be maximum simultaneously when total mechanical energy is constant. At the extremes, speed is zero, so kinetic energy is zero and potential energy is maximum."},{"text":"

Kinetic energy is zero
Potential energy is minimum

","correct":false,"explanation":"Although kinetic energy is zero at an extreme, potential energy is not minimum. Maximum displacement gives maximum potential energy in simple harmonic motion."}]},{"id":"49adbecf-41b8-4eee-afe2-25b035bacb75","cardType":"basic","points":2,"front":"Describe the energy stores when an oscillator passes through the equilibrium position ($x = 0$).","back":"

Kinetic energy is maximum
Potential energy is minimum

","mcqOptions":[{"text":"

Kinetic energy is maximum
Potential energy is minimum

","correct":true,"explanation":"At equilibrium, displacement and potential energy are minimum while speed is maximum. Consequently, the oscillator’s kinetic energy is maximum."},{"text":"

Kinetic energy is zero
Potential energy is maximum

","correct":false,"explanation":"This describes an extreme position rather than equilibrium. At equilibrium, the oscillator has maximum speed, so kinetic energy is maximum and potential energy is minimum."},{"text":"

Kinetic energy is minimum
Potential energy is minimum

","correct":false,"explanation":"Total mechanical energy remains constant, so both stores cannot be minimum together. At equilibrium, potential energy is minimum but kinetic energy is maximum."},{"text":"

Kinetic energy is maximum
Potential energy is maximum

","correct":false,"explanation":"This would make the total energy exceed its constant value. At equilibrium, maximum speed gives maximum kinetic energy, while zero displacement gives minimum potential energy."}]},{"id":"474d1276-40f3-42a5-a4e7-e533608887e0","cardType":"cloze","points":null,"front":"In undamped simple harmonic motion, the total mechanical energy remains [ ].","back":"In undamped simple harmonic motion, the total mechanical energy remains constant.","mcqOptions":[{"text":"constant","correct":true,"explanation":"With no damping, no mechanical energy is transferred to the surroundings. Energy continually shifts between kinetic and potential stores, but their total remains constant."},{"text":"decreasing","correct":false,"explanation":"A decreasing total energy describes damped motion, where resistive forces transfer energy to the surroundings. In undamped simple harmonic motion, total mechanical energy remains constant."},{"text":"increasing","correct":false,"explanation":"Total energy would increase only if an external force supplied energy. Undamped simple harmonic motion conserves mechanical energy, so the total remains constant."},{"text":"variable","correct":false,"explanation":"The kinetic and potential energies vary individually, which may make this tempting. Their changes compensate exactly, so total mechanical energy remains constant."}]},{"id":"378966fd-3c6b-40ff-a7a5-7084c49db8e1","cardType":"basic","points":1,"front":"What is the formula for the elastic potential energy ($E_p$) of a spring-mass system at displacement $x$?","back":"$$E_p = \\frac{1}{2}kx^2$","mcqOptions":[{"text":"$$E_p = \\frac{1}{2}kx^2$","correct":true,"explanation":"Spring force increases linearly from zero to $kx$ during extension. The area under the force-extension graph is therefore $\\frac{1}{2}x(kx) = \\frac{1}{2}kx^2$."},{"text":"$$E_p = \\frac{1}{2}mv^2$","correct":false,"explanation":"This is the formula for kinetic energy, not elastic potential energy. A spring displaced by $x$ stores $E_p = \\frac{1}{2}kx^2$."},{"text":"$$E_p = kx^2$","correct":false,"explanation":"This misses the factor of one-half obtained from the area under the force-extension graph. The correct elastic potential energy is $E_p = \\frac{1}{2}kx^2$."},{"text":"$$E_p = \\frac{1}{2}kx$","correct":false,"explanation":"This fails to square the displacement and is dimensionally a force rather than an energy. Elastic potential energy is $E_p = \\frac{1}{2}kx^2$."}]},{"id":"899345c1-9ae3-44a5-a95c-e6d57e508f6f","cardType":"basic","points":1,"front":"How is the total mechanical energy ($E_{total}$) calculated for a spring-mass system using amplitude $A$?","back":"$$E_{total} = \\frac{1}{2}kA^2$","mcqOptions":[{"text":"$$E_{total} = \\frac{1}{2}kA^2$","correct":true,"explanation":"At maximum displacement $A$, speed and kinetic energy are zero, so all the mechanical energy is elastic potential energy: $E_{total} = \\frac{1}{2}kA^2$."},{"text":"$$E_{total} = \\frac{1}{2}kx^2$","correct":false,"explanation":"This gives the potential energy at a general displacement $x$, not the constant total energy. Using the maximum displacement gives $E_{total} = \\frac{1}{2}kA^2$."},{"text":"$$E_{total} = kA^2$","correct":false,"explanation":"This omits the factor of one-half from the elastic potential energy formula, doubling the result. The total energy is $E_{total} = \\frac{1}{2}kA^2$."},{"text":"$$E_{total} = \\frac{1}{2}kA$","correct":false,"explanation":"This treats energy as proportional to amplitude rather than amplitude squared and is dimensionally incorrect. The total mechanical energy is $E_{total} = \\frac{1}{2}kA^2$."}]},{"id":"ce0847b8-f12e-4b95-aa88-e72e1cbd7f0c","cardType":"basic","points":1,"front":"What is the formula for kinetic energy ($E_k$) of a spring-mass system at a specific displacement $x$?","back":"$$E_k = \\frac{1}{2}k(A^2 - x^2)$","mcqOptions":[{"text":"$$E_k = \\frac{1}{2}k(A^2 - x^2)$","correct":true,"explanation":"Kinetic energy equals total mechanical energy minus elastic potential energy. Subtracting $\\frac{1}{2}kx^2$ from $\\frac{1}{2}kA^2$ gives this expression."},{"text":"$$E_k = \\frac{1}{2}k(A^2 + x^2)$","correct":false,"explanation":"This adds potential energy to the total instead of subtracting it. Since $E_k = E_{total} - E_p$, the result is $E_k = \\frac{1}{2}k(A^2 - x^2)$."},{"text":"$$E_k = \\frac{1}{2}kx^2$","correct":false,"explanation":"This is the elastic potential energy at displacement $x$. Kinetic energy is the remaining mechanical energy, $E_k = \\frac{1}{2}k(A^2 - x^2)$."},{"text":"$$E_k = \\frac{1}{2}k(A - x)^2$","correct":false,"explanation":"This subtracts the displacements before squaring, introducing an incorrect cross-term. The squared quantities must be subtracted: $E_k = \\frac{1}{2}k(A^2 - x^2)$."}]},{"id":"33a6a1bb-8b4e-4430-a724-3ddea020e311","cardType":"cloze","points":null,"front":"Because total energy is proportional to amplitude squared ($E \\propto A^2$), tripling the amplitude increases the total energy by a factor of [ ].","back":"Because total energy is proportional to amplitude squared ($E \\propto A^2$), tripling the amplitude increases the total energy by a factor of 9.","mcqOptions":[{"text":"9","correct":true,"explanation":"Total energy is proportional to the square of amplitude. Replacing $A$ with $3A$ multiplies the energy by $3^2$, giving a factor of 9."},{"text":"3","correct":false,"explanation":"This treats energy as directly proportional to amplitude. Because energy depends on amplitude squared, tripling amplitude multiplies energy by $3^2 = 9$."},{"text":"27","correct":false,"explanation":"This cubes the amplitude factor rather than squaring it. Since $E \\propto A^2$, tripling $A$ increases the total energy by $3^2 = 9$."},{"text":"6","correct":false,"explanation":"This incorrectly doubles the amplitude multiplier rather than squaring it. The relationship $E \\propto A^2$ means a tripled amplitude produces an energy factor of $3^2 = 9$."}]},{"id":"b9f7f616-2afc-4e71-ac5d-06bde3792759","cardType":"basic","points":1,"front":"State the general equation for total energy in SHM using mass ($m$) and angular frequency ($\\omega$).","back":"$$E_{total} = \\frac{1}{2}m\\omega^2A^2$","mcqOptions":[{"text":"$$E_{total} = \\frac{1}{2}m\\omega^2A^2$","correct":true,"explanation":"At maximum displacement, all the oscillator’s energy is potential. Substituting $k=m\\omega^2$ into $E=\\frac{1}{2}kA^2$ gives this general expression."},{"text":"$$E_{total} = \\frac{1}{2}m\\omega A^2$","correct":false,"explanation":"This misses the square on angular frequency, making the dimensions incorrect. Using $k=m\\omega^2$ in $E=\\frac{1}{2}kA^2$ gives $E_{total}=\\frac{1}{2}m\\omega^2A^2$."},{"text":"$$E_{total} = \\frac{1}{2}m\\omega^2A$","correct":false,"explanation":"This uses amplitude rather than amplitude squared. SHM energy is proportional to $A^2$, so the correct equation is $E_{total}=\\frac{1}{2}m\\omega^2A^2$."},{"text":"$$E_{total} = m\\omega^2A^2$","correct":false,"explanation":"This omits the factor of one-half from the elastic potential energy equation. The correct total energy is $E_{total}=\\frac{1}{2}m\\omega^2A^2$."}]},{"id":"48057413-262b-4185-8932-b6f8a75e634e","cardType":"cloze","points":null,"front":"In a spring-mass oscillator, the relationship between the spring constant ($k$), mass ($m$), and angular frequency ($\\omega$) is $\\text{[...]}$.","back":"In a spring-mass oscillator, the relationship between the spring constant ($k$), mass ($m$), and angular frequency ($\\omega$) is $\\boxed{k = m\\omega^2}$.","mcqOptions":[{"text":"$$k = m\\omega^2$","correct":true,"explanation":"For a spring-mass oscillator, $\\omega=\\sqrt{k/m}$. Squaring and rearranging this relationship gives $k=m\\omega^2$."},{"text":"$$k = \\frac{m}{\\omega^2}$","correct":false,"explanation":"This reverses the dependence on angular frequency. Since $\\omega=\\sqrt{k/m}$, rearrangement gives $k=m\\omega^2$, so greater stiffness produces a greater angular frequency."},{"text":"$$k = m\\omega$","correct":false,"explanation":"This misses the square on angular frequency and is dimensionally inconsistent. Squaring $\\omega=\\sqrt{k/m}$ gives $\\omega^2=k/m$, hence $k=m\\omega^2$."},{"text":"$$k = \\frac{\\omega^2}{m}$","correct":false,"explanation":"This incorrectly divides by mass when rearranging the angular-frequency equation. From $\\omega^2=k/m$, multiplying by $m$ gives the correct relationship $k=m\\omega^2$."}]},{"id":"7148e364-a7f4-431d-a30f-62f68c34b049","cardType":"basic","points":1,"front":"Why are the kinetic and potential energies in SHM never negative?","back":"They depend on squared terms (displacement $x^2$ or velocity $v^2$).","mcqOptions":[{"text":"They depend on squared terms (displacement $x^2$ or velocity $v^2$).","correct":true,"explanation":"Both $x^2$ and $v^2$ are non-negative for real displacement and velocity. Therefore $E_p=\\frac{1}{2}kx^2$ and $E_k=\\frac{1}{2}mv^2$ cannot be negative."},{"text":"They depend on absolute terms (displacement $|x|$ or velocity $|v|$).","correct":false,"explanation":"SHM energy equations use squares, not absolute values. The expressions $E_p=\\frac{1}{2}kx^2$ and $E_k=\\frac{1}{2}mv^2$ are non-negative because $x^2$ and $v^2$ cannot be negative."},{"text":"They depend on positive directions (displacement $x$ or velocity $v$).","correct":false,"explanation":"Displacement and velocity can each be positive or negative depending on direction. Energy remains non-negative because the equations contain the squared quantities $x^2$ and $v^2$."},{"text":"They depend on constant terms (spring constant $k$ or mass $m$).","correct":false,"explanation":"Positive values of $k$ and $m$ alone do not explain the directional independence. The crucial feature is that displacement and velocity are squared, making both energy stores non-negative."}]},{"id":"67f2d45e-ce1d-4f0d-90ab-efa452dbd498","cardType":"cloze","points":null,"front":"If the displacement of an oscillator follows $x = A\\cos(\\omega t)$, the potential energy varies according to $\\text{[...]}$.","back":"If the displacement of an oscillator follows $x = A\\cos(\\omega t)$, the potential energy varies according to $\\boxed{E_p = E_{total}\\cos^2(\\omega t)}$.","mcqOptions":[{"text":"$$E_p = E_{total}\\cos^2(\\omega t)$","correct":true,"explanation":"Potential energy is proportional to $x^2$. Substituting $x=A\\cos(\\omega t)$ into $E_p=\\frac{1}{2}m\\omega^2x^2$ gives $E_p=E_{total}\\cos^2(\\omega t)$."},{"text":"$$E_p = E_{total}\\cos(\\omega t)$","correct":false,"explanation":"This fails to square the displacement dependence and would allow negative potential energy. Since $E_p\\propto x^2$, the correct variation is $E_p=E_{total}\\cos^2(\\omega t)$."},{"text":"$$E_p = E_{total}\\sin^2(\\omega t)$","correct":false,"explanation":"This is the kinetic-energy variation for the stated cosine displacement. Potential energy is greatest when $|x|=A$, so it varies as $E_p=E_{total}\\cos^2(\\omega t)$."},{"text":"$$E_p = E_{total}\\cos^2(2\\omega t)$","correct":false,"explanation":"Squaring the cosine already makes the energy repeat twice per oscillation; its argument remains $\\omega t$. The correct expression is $E_p=E_{total}\\cos^2(\\omega t)$."}]},{"id":"3973acf0-46cf-42bc-84d7-7feb3873bbcb","cardType":"basic","points":1,"front":"How many times does the kinetic energy reach a maximum during one complete oscillation period ($T$)?","back":"Twice (at each pass through equilibrium).","mcqOptions":[{"text":"Twice (at each pass through equilibrium).","correct":true,"explanation":"The oscillator passes through equilibrium twice per complete cycle. Its speed, and therefore its kinetic energy, reaches a maximum during each of these passes."},{"text":"Once (at the positive maximum displacement).","correct":false,"explanation":"At maximum displacement the oscillator is momentarily stationary, so its kinetic energy is zero. Kinetic energy instead reaches a maximum twice, at each pass through equilibrium."},{"text":"Twice (at each maximum displacement).","correct":false,"explanation":"Although there are two maximum-displacement positions, the oscillator is stationary at both, so kinetic energy is zero there. It reaches a maximum twice at equilibrium."},{"text":"Four times (at each quarter of the period).","correct":false,"explanation":"Dividing the period into quarters overcounts the maxima. The oscillator crosses equilibrium only twice per period, and kinetic energy is maximal at those two crossings."}]},{"id":"77ae0c3b-98f7-43ca-b8fd-8a16dcc42983","cardType":"basic","points":1,"front":"If an oscillator has a frequency $f$, what is the frequency of the energy variations ($f_E$)?","back":"$$f_E = 2f$","mcqOptions":[{"text":"$$f_E = 2f$","correct":true,"explanation":"Energy depends on squared displacement or velocity, so it has the same value after half an oscillation period. Its period is $T/2$, giving $f_E=2f$."},{"text":"$$f_E = f$","correct":false,"explanation":"This assumes energy repeats only after a full oscillation. Squared displacement and velocity repeat after $T/2$, so the energy frequency is twice the oscillation frequency: $f_E=2f$."},{"text":"$$f_E = \\frac{f}{2}$","correct":false,"explanation":"This incorrectly doubles the energy period rather than halving it. Energy repeats every $T/2$, so its frequency is $1/(T/2)=2f$."},{"text":"$$f_E = 4f$","correct":false,"explanation":"This counts four quarter-cycle stages as four complete energy cycles. Each energy graph completes two cycles per oscillation period, so the correct frequency is $f_E=2f$."}]},{"id":"0b77dfed-fc13-4789-a916-9277def7c056","cardType":"cloze","points":null,"front":"The kinetic and potential energy graphs repeat every $\\text{[...]}$ seconds, where $T$ is the period of the oscillation.","back":"The kinetic and potential energy graphs repeat every $\\boxed{\\frac{T}{2}}$ seconds, where $T$ is the period of the oscillation.","mcqOptions":[{"text":"$$\\frac{T}{2}$","correct":true,"explanation":"Kinetic and potential energies depend on squared sinusoidal quantities. Squaring removes the sign change after half a cycle, so both energy graphs repeat every $T/2$ seconds."},{"text":"$$T$","correct":false,"explanation":"This is the period of displacement, not energy. Because energy depends on squared displacement or velocity, its values repeat after half an oscillation period, every $T/2$ seconds."},{"text":"$$2T$","correct":false,"explanation":"This incorrectly assumes energy varies more slowly than displacement. Squared sinusoidal quantities repeat after half the displacement period, so the energy graphs repeat every $T/2$ seconds."},{"text":"$$\\frac{T}{4}$","correct":false,"explanation":"Quarter-period intervals mark changes between maximum and minimum energy, not a full repetition of an energy graph. The complete energy pattern repeats every $T/2$ seconds."}]},{"id":"dfb224ad-f222-444c-82c0-c1f53d7da9fc","cardType":"cloze","points":null,"front":"For a simple pendulum, the potential energy store is primarily [ ].","back":"For a simple pendulum, the potential energy store is primarily gravitational potential energy.","mcqOptions":[{"text":"gravitational potential energy","correct":true,"explanation":"As a pendulum bob rises and falls, its height in Earth’s gravitational field changes. Its potential energy store is therefore primarily gravitational potential energy."},{"text":"elastic potential energy","correct":false,"explanation":"Elastic potential energy requires deformation of an elastic object, such as a spring. A simple pendulum changes height instead, so its potential energy store is gravitational potential energy."},{"text":"electric potential energy","correct":false,"explanation":"Electric potential energy concerns charges in an electric field, which is not the restoring mechanism of a simple pendulum. Its changing height gives it gravitational potential energy."},{"text":"chemical potential energy","correct":false,"explanation":"Chemical potential energy is associated with chemical bonds and reactions, not the pendulum’s position. The bob’s changing height means its relevant store is gravitational potential energy."}]},{"id":"cb37e805-eb62-4c74-ae7f-ad5ee34e47e5","cardType":"basic","points":2,"front":"Define damping in terms of energy and forces.","back":"

Resistive forces (e.g. friction) act against motion
Mechanical energy is transferred to thermal energy

","mcqOptions":[{"text":"

Resistive forces (e.g. friction) act against motion
Mechanical energy is transferred to thermal energy

","correct":true,"explanation":"Damping arises when resistive forces oppose an oscillator’s motion and transfer its mechanical energy to the surroundings, usually as thermal energy."},{"text":"

Restoring forces act towards equilibrium
Kinetic energy is transferred to potential energy

","correct":false,"explanation":"Restoring forces produce oscillation rather than damping, while the exchange between kinetic and potential energy conserves mechanical energy. Damping requires resistive forces that transfer mechanical energy to thermal energy."},{"text":"

Driving forces act in the direction of motion
External energy is transferred to mechanical energy

","correct":false,"explanation":"A driving force supplies energy to an oscillator and can maintain or increase its amplitude. Damping instead involves resistive forces opposing motion and transferring mechanical energy to thermal energy."},{"text":"

Resistive forces (e.g. friction) act against motion
Thermal energy is transferred to mechanical energy

","correct":false,"explanation":"The direction of energy transfer has been reversed. Resistive forces remove mechanical energy from the oscillator and transfer it to thermal energy, rather than converting thermal energy into mechanical energy."}]},{"id":"44d9f3bf-39d4-45c0-9cb5-257b3bd90762","cardType":"basic","points":1,"front":"What is the physical effect of damping on an oscillator's amplitude over time?","back":"The amplitude decreases.","mcqOptions":[{"text":"The amplitude decreases.","correct":true,"explanation":"Damping continuously removes mechanical energy from the oscillator, so its maximum displacement from equilibrium, or amplitude, becomes progressively smaller with time."},{"text":"The amplitude increases.","correct":false,"explanation":"An increasing amplitude indicates that energy is being supplied, as in driven oscillation. Damping removes mechanical energy, so the oscillator’s amplitude decreases over time."},{"text":"The amplitude remains constant.","correct":false,"explanation":"A constant amplitude describes an undamped oscillator or one receiving enough driving energy to replace losses. With damping alone, energy is lost and the amplitude decreases."},{"text":"The amplitude decreases then increases.","correct":false,"explanation":"Passive damping does not later restore the oscillator’s lost mechanical energy. The amplitude envelope continues to decrease rather than decreasing and then increasing."}]},{"id":"0f3be3d1-47b8-454d-ab63-a5130581056c","cardType":"cloze","points":null,"front":"Simple harmonic motion (SHM) is defined by the acceleration equation $\\text{[...]}$.","back":"Simple harmonic motion (SHM) is defined by the acceleration equation $\\boxed{a = -\\omega^2x}$.","mcqOptions":[{"text":"$$a = -\\omega^2x$","correct":true,"explanation":"In SHM, acceleration is proportional to displacement and directed towards equilibrium. The minus sign gives the opposite direction, while the constant of proportionality is $\\omega^2$."},{"text":"$$a = \\omega^2x$","correct":false,"explanation":"Omitting the minus sign makes acceleration point away from equilibrium, producing unstable motion rather than SHM. The correct defining equation is $a = -\\omega^2x$."},{"text":"$$a = -\\omega x$","correct":false,"explanation":"Using $\\omega$ rather than $\\omega^2$ is dimensionally inconsistent because $\\omega x$ has units of velocity. SHM requires $a = -\\omega^2x$."},{"text":"$$a = -\\omega^2v$","correct":false,"explanation":"SHM acceleration is proportional to displacement, not velocity. Acceleration proportional to negative velocity describes a damping contribution, whereas the defining SHM equation is $a = -\\omega^2x$."}]},{"id":"3c3edb55-09bb-4af4-8d01-e996b34e1d5e","cardType":"basic","points":2,"front":"How do you calculate the energy dissipated by damping if the initial and final amplitudes are known?","back":"Calculate $\\Delta E = E_{initial} - E_{final}$ using the $E \\propto A^2$ relationship.","mcqOptions":[{"text":"Calculate $\\Delta E = E_{initial} - E_{final}$ using the $E \\propto A^2$ relationship.","correct":true,"explanation":"Damping removes the difference between the initial and final mechanical energies. Since oscillator energy is proportional to amplitude squared, each energy must be calculated using the corresponding squared amplitude."},{"text":"Calculate $\\Delta E = E_{final} - E_{initial}$ using the $E \\propto A^2$ relationship.","correct":false,"explanation":"Reversing the subtraction gives a negative value when damping has reduced the energy. The positive energy dissipated is $\\Delta E = E_{initial} - E_{final}$, with $E \\propto A^2$."},{"text":"Calculate $\\Delta E = E_{initial} - E_{final}$ using the $E \\propto A$ relationship.","correct":false,"explanation":"Oscillator energy is proportional to the square of amplitude, not amplitude itself. The correct method uses $E \\propto A^2$ and subtracts final energy from initial energy."},{"text":"Calculate $\\Delta E = E_{initial} + E_{final}$ using the $E \\propto A^2$ relationship.","correct":false,"explanation":"Adding the initial and final energies does not measure the energy lost between the two states. Energy dissipated is their difference, $E_{initial} - E_{final}$, using $E \\propto A^2$."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Energy in oscillations","title":"Energy in oscillations Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"e171d40a-7075-5308-b145-1ef1c8a53e88","walled":"$L39"}],"$L3a","$L3b"]