Analyze Wave Amplitude and Energy - Grade 11 Physics

A monochromatic electromagnetic beam illuminates a fixed area. Its electric-field amplitude is doubled while its frequency is reduced to one-half of its original value. Assuming the beam remains steady, how does the number of photons arriving each second change?
Answer: It increases by a factor of eight, because the beam power quadruples while the energy carried by each photon is halved Whenever a question links electromagnetic wave properties to photon counts, you need to track two separate effects: how the wave's intensity changes, and how each photon's energy changes — then combine them. Start with intensity. The intensity of an electromagnetic wave is proportional to the square of the electric-field amplitude: $I \propto E_0^2$ . Doubling the amplitude means $I \propto (2E_0)^2 = 4E_0^2$ , so the beam's power delivered to the fixed area quadruples. Now consider photon energy. Each photon carries energy $E_{photon} = hf$ . Halving the frequency means each photon now carries only half the energy it did before. The number of photons arriving per second is power divided by energy per photon: $N = \frac{P}{hf}$ Combining both changes: power goes up by 4, and energy per photon goes down by 2, so: $N_{new} = \frac{4P}{\frac{1}{2}hf} = 8 \cdot \frac{P}{hf} = 8N$ The photon rate increases by a factor of eight, confirming answer C. Answer A is wrong because it only accounts for the frequency change and ignores the quadrupled power — both factors must be included. Answer B is wrong because it correctly identifies the quadrupled power but forgets that lower-frequency photons are "cheaper" in energy, so each watt of power delivers more photons, not the same number. Answer D reverses the logic entirely — lower photon energy actually increases the photon rate for a given power, not decreases it. Your strategy: always split these problems into "how does power change?" and "how does energy-per-photon change?" then divide to find the photon rate.
Doubling the electric-field amplitude of an EM wave changes the energy per photon by:
Answer: No change An EM wave's electric-field amplitude determines the wave's intensity, not the energy of a single photon. Photon energy is E = hf, set by frequency alone, so doubling the electric-field amplitude leaves the energy per photon unchanged. The tempting wrong answer is 4 times, because intensity is proportional to amplitude squared, but that affects how many photons are present, not how much energy each photon carries.
For a sinusoidal wave on a string, the average power is proportional to $A^2f^2v$ . A second wave has half the amplitude and twice the frequency of the first wave. The string tension is also quadrupled without changing the string's linear density. How does the second wave's average power compare with that of the first?
Answer: It is twice as great because the speed doubles after other factors cancel Whenever you see a wave power problem with multiple changing quantities, your strategy should be to track each variable's effect on the power formula separately, then combine them multiplicatively. The average power formula is $P \propto A^2 f^2 v$ . Let's apply the given changes one at a time. The second wave has half the amplitude, so its $A^2$ contribution becomes $(A/2)^2 = A^2/4$ , reducing power by a factor of 4. The frequency doubles, so $f^2$ becomes $(2f)^2 = 4f^2$ , increasing power by a factor of 4. These two effects exactly cancel — but we're not done. The tension is quadrupled while linear density stays the same. Since wave speed on a string is $v = \sqrt{T/\mu}$ , quadrupling tension gives $v' = \sqrt{4T/\mu} = 2v$ . This doubles the speed, and therefore doubles the power. The net result: $\frac{1}{4} \times 4 \times 2 = 2$ , so the second wave carries twice the power. That confirms D. Choice A is tempting because the amplitude and frequency changes do cancel each other — but ignoring the tension change means you've left the problem half-finished. Choice B correctly recognizes that frequency and tension both increase the power, but fails to account for the amplitude reduction, which cuts power by a factor of 4. Choice C overstates the amplitude's dominance; it cancels with the frequency term rather than dominating it. A useful habit: list every changed quantity, evaluate its exponent in the formula, and only then multiply all the factors together — never assume two effects are the whole story.
With the same frequency and medium, wave A delivers 4 times B's power. A has amplitude 6 cm. B's amplitude is:
Answer: 3.0 cm Since power is proportional to amplitude squared, four times the power means only twice the amplitude. Wave A at 6 cm is twice wave B's amplitude, so B must be 3.0 cm. The common error is 1.5 cm, which treats power as directly proportional to amplitude instead of amplitude squared.
With the same frequency and medium, a 2.0-cm-amplitude wave carries 10 J/m. An 8.0-cm-amplitude wave carries:
Answer: 160 J/m Wave energy is proportional to amplitude squared. Increasing amplitude from 2.0 cm to 8.0 cm multiplies it by 4, so the energy multiplies by $4^2$ = 16. Thus 10 J/m times 16 gives 160 J/m. The tempting error is using a direct 4-times multiplier and getting 40 J/m, but that forgets the squared dependence.
Two pulses with the same general shape travel on the same string. Pulse B has $0.80$ times the amplitude of pulse A but is $2.5$ times as wide along the string. Assuming pulse energy is proportional to amplitude squared and spatial width, how does the energy of pulse B compare with that of pulse A?
Answer: It is $1.6$ times as large because both amplitude squared and width contribute When a question gives you a formula built from multiple variables, your job is to apply it carefully to each variable separately, then combine the results — don't let one variable overshadow the other. Here, the energy of a pulse is proportional to $A^2 \times w$ , where $A$ is amplitude and $w$ is width. Think of this as two separate factors multiplying together. For pulse B, the amplitude is $0.80A$ , so the amplitude-squared factor becomes $(0.80)^2 = 0.64$ times that of pulse A. The width factor is $2.5$ times that of pulse A. Multiplying both contributions together: $E_B = 0.64 \times 2.5 \times E_A = 1.6 \times E_A$ . Pulse B carries 1.6 times the energy of pulse A, confirming that D is correct. Choice A makes the classic mistake of ignoring width entirely — it correctly calculates $(0.80)^2 = 0.64$ but stops there, as if width plays no role in energy. Choice B assumes both factors contribute linearly, computing $0.80 \times 2.5 = 2.0$ — but amplitude enters as a square, not a first power, so this underestimates how much the smaller amplitude reduces energy. Choice C goes to the opposite extreme from A, discarding amplitude altogether and keeping only the width factor of $2.5$ . A useful strategy: whenever a formula has multiple variables, calculate each factor's scaling ratio independently, then multiply them together at the end. This prevents you from accidentally dropping a term or forgetting an exponent — both of which are traps this question is specifically designed to catch.
To double the energy a wave transports, multiply its amplitude by:
Answer: √2 times Wave energy goes as amplitude squared, so multiplying amplitude by 2 makes energy 4 times larger. To get exactly twice the energy, you need the square of the new amplitude to be twice the old, which means multiplying by the square root of 2. The tempting wrong answer is 2 times, but that quadruples energy instead of doubling it.
A wave's amplitude is increased by 30%. Its energy transport rate increases by:
Answer: 69% Amplitude becomes 1.30 times its original value. Energy transport rate is proportional to amplitude squared, so multiply the factor by itself: 1.30 x 1.30 = 1.69. That means a 69% increase. The tempting 30% answer treats energy as directly proportional to amplitude, but it actually varies with the square.
A generator produces a pulse on a uniform string. A second pulse has the same general shape and travels on the same string, but its amplitude is $1.5$ times as large and the generator acts for only $0.40$ times as long. Assuming pulse energy is proportional to amplitude squared and to pulse duration, what is the ratio of the second pulse's energy to the first pulse's energy?
Answer: $0.90$ , because both amplitude squared and duration must be included When a question gives you a proportionality relationship and asks you to compare two scenarios, your job is to carefully apply every factor in that relationship — no more, no less. Here, you're told that pulse energy satisfies $E \propto A^2 \cdot t$ , where $A$ is amplitude and $t$ is duration. To find the ratio of the second pulse's energy to the first, substitute the given changes: the amplitude scales by $1.5$ and the duration scales by $0.40$ . This gives: $\frac{E_2}{E_1} = \frac{(1.5A)^2 \cdot (0.40t)}{A^2 \cdot t} = (1.5)^2 \times 0.40 = 2.25 \times 0.40 = 0.90$ So the correct answer is B. Choice A claims amplitude affects energy linearly, computing $1.5 \times 0.40 = 0.60$ . This ignores the squared relationship for amplitude — a critical part of the given formula. Choice C gets $(1.5)^2 = 2.25$ right for the amplitude term but completely ignores the duration factor. Dropping one of the proportional factors is a classic partial-credit trap. Choice D squares the entire product $1.5 \times 0.40 = 0.60$ , giving $0.36$ . This misapplies the exponent — only the amplitude is squared, not the duration, and certainly not their product. Study tip: Whenever a problem gives you a proportionality rule, write it out explicitly before plugging in numbers. Identify which variables carry exponents and apply them individually — don't distribute exponents across the whole expression unless the math truly calls for it.
A wave pulse initially carries $12~\text{J}$ of energy. It passes through two identical damping sections, each of which reduces the pulse's amplitude to $75\%$ of the amplitude entering that section without noticeably changing its shape or width. How much energy remains after both sections?
Answer: Approximately $3.8~\text{J}$ , because the combined amplitude ratio is $(0.75)^2$ and energy is proportional to amplitude squared Whenever you see a wave problem involving damping, the key relationship to recall is that wave energy is proportional to the square of the amplitude: $E \propto A^2$ . This single fact unlocks the entire problem. Each damping section reduces the amplitude by a factor of $0.75$ . After passing through both sections, the combined amplitude ratio is $(0.75)^2 = 0.5625$ . Because energy scales with amplitude squared, the energy ratio becomes $(0.75)^2)^2 = (0.75)^4$ . Wait — let's be careful. Each section multiplies the amplitude by $0.75$ , so the energy after each section is multiplied by $(0.75)^2 = 0.5625$ . After two sections, the total energy factor is $(0.5625)^2 = (0.75)^4 \approx 0.3164$ . Multiplying: $12 \times 0.3164 \approx 3.8~\text{J}$ . Answer A is correct. Answer B claims the energy ratio equals $(0.75)^2$ directly, confusing the amplitude ratio with the energy ratio — it forgets to square again for the second section, giving $12 \times 0.5625 \approx 6.8~\text{J}$ . This is the most tempting trap because it correctly squares once but stops too early. Answer C adds the reductions ( $0.75 + 0.75 = 1.5$ , then interprets a "50% total loss"), which is mathematically invalid — ratios must be multiplied, not added. Answer D applies only a single amplitude reduction, ignoring the second damping section entirely. Your study tip: always track whether a quantity scales with $A$ or $A^2$ . When amplitude is reduced by a factor $f$ , energy is reduced by $f^2$ — apply this per section, then multiply the results together.
A radio transmitter radiates uniformly in all directions. Its power output is increased to four times its original value while a receiver is moved to three times its original distance from the transmitter. What is the ratio of the new electric-field amplitude at the receiver to the original amplitude?
Answer: $2/3$ , because field amplitude depends on the square root of intensity Whenever you see a question linking power output, distance, and electric-field amplitude, you need to track two separate relationships carefully before combining them. Start with intensity. Since the transmitter radiates uniformly in all directions, intensity spreads over a sphere: $I = \frac{P}{4\pi r^2}$ . This means intensity is proportional to $\frac{P}{r^2}$ . When power quadruples and distance triples, the new intensity is: $I_{new} = I_{old} \times \frac{4}{3^2} = \frac{4}{9}\, I_{old}$ Now here's the critical step: electric-field amplitude $E$ is not directly proportional to intensity. The relationship is $I \propto E^2$ , which means $E \propto \sqrt{I}$ . So to find the ratio of field amplitudes: $\frac{E_{new}}{E_{old}} = \sqrt{\frac{I_{new}}{I_{old}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$ That confirms C is correct. Choice A is wrong because it ignores the increased power entirely — both changes must be accounted for. Choice B gives you the correct intensity ratio ( $4/9$ ), but stops one step too early; the question asks about electric-field amplitude, not intensity, and those two quantities are not the same thing. Choice D correctly notes that quadrupling power doubles the field amplitude ( $\sqrt{4} = 2$ ), but completely ignores the tripling of distance, which introduces a factor of $1/3$ . A useful strategy: on problems like this, write out $E \propto \sqrt{P}/r$ directly. This single expression shows that field amplitude scales with the square root of power and inversely with distance — keeping both effects visible so you never drop one by accident.
At a fixed location, a sound level rises from $68~\text{dB}$ to $80~\text{dB}$ . The frequency and properties of the air remain unchanged. Approximately how many times greater is the pressure amplitude of the louder sound?
Answer: $4.0$ times greater, after converting the decibel increase to an intensity ratio and then taking a square root Whenever a question links decibels to physical wave properties like pressure amplitude, you need a two-step conversion — decibels don't map directly to pressure. Start by converting the decibel change to an intensity ratio. The decibel formula is $\Delta L = 10\log_{10}(I_2/I_1)$ . Here, $\Delta L = 80 - 68 = 12~\text{dB}$ , so: $\frac{I_2}{I_1} = 10^{12/10} = 10^{1.2} \approx 15.8$ The intensity is about 15.8 times greater. But the question asks about pressure amplitude, not intensity. Since intensity is proportional to the square of pressure amplitude ( $I \propto p^2$ ), you take the square root: $\frac{p_2}{p_1} = \sqrt{15.8} \approx 4.0$ That confirms B is correct. Now for the traps. A is wrong because dividing the decibel increase by ten (getting 1.2) misinterprets the logarithmic scale — 1.2 is just the exponent, not a meaningful ratio on its own. C is wrong because it treats the 12 dB rise as a direct multiplier of pressure amplitude, ignoring both the logarithmic nature of decibels and the square-root relationship between pressure and intensity. D is the sneaky one — 15.8 is the correct intensity ratio, but the question specifically asks for pressure amplitude, and skipping the square root step gives you the wrong physical quantity. Your takeaway: on any decibel problem asking about pressure, always do two steps — convert dB to intensity ratio using $10^{\Delta L/10}$ , then take the square root for pressure amplitude.
Two identical waves, each with amplitude $A$ and average power $P$ , travel in opposite directions on a lossless string and form a standing wave. At an antinode, the displacement amplitude is $2A$ . Which statement correctly relates this amplitude to energy transfer?
Answer: The net average power is zero because the component waves transport equal energy oppositely Whenever you see a question about standing waves, the key distinction to make is between displacement and energy transport. A standing wave forms when two identical waves travel in opposite directions — and this superposition has dramatic consequences for power flow. Each component wave carries average power $P$ in its direction of travel. Because the string is lossless and the waves are identical, one wave transports energy to the right at rate $P$ while the other transports energy to the left at rate $P$ . The net average power at any point along the string is therefore $P - P = 0$ . This is the defining energetic signature of a true standing wave: energy sloshes back and forth between kinetic and potential forms, but no net energy is transmitted past any fixed point. Answer B is correct. Answer A is tempting because the antinode amplitude $2A$ is double the individual amplitude, and since power scales as amplitude squared, you might expect $4P$ . But that reasoning applies to a single traveling wave — it does not account for the fact that two waves are traveling in opposite directions, canceling each other's net transport. Answer C makes the same mistake by simply adding the two powers without considering their opposing directions. Answer D is also wrong: opposite displacements do sometimes occur (at nodes), but the reason power is zero is not partial cancellation of displacement — it is the complete cancellation of energy flux from two equally powerful, counter-propagating waves. A useful rule to memorize: standing wave = zero net power transfer, regardless of how large the antinode amplitude appears. Amplitude tells you about local displacement, not about energy flow direction.
Two coherent waves of equal individual amplitude reach the same detector. Their phase difference initially is $60°$ and later becomes $120°$ , while each wave's individual amplitude remains unchanged. What is the ratio of the later detected intensity to the initial detected intensity?
Answer: $1/3$ , because the resultant amplitude must be squared when comparing intensities When two coherent waves interfere, the resultant amplitude depends on their phase difference through the formula $A_{res} = 2A\cos(\delta/2)$ , where $A$ is each wave's individual amplitude and $\delta$ is the phase difference. Since intensity is proportional to the square of amplitude ( $I \propto A_{res}^2$ ), you must always square the resultant amplitude when comparing intensities — this is the key idea being tested here. For the initial phase difference of $60°$ : $A_{res,i} = 2A\cos(30°) = 2A \cdot \frac{\sqrt{3}}{2} = A\sqrt{3}$ , giving $I_i \propto 3A^2$ . For the later phase difference of $120°$ : $A_{res,f} = 2A\cos(60°) = 2A \cdot \frac{1}{2} = A$ , giving $I_f \propto A^2$ . The ratio is $I_f/I_i = A^2/3A^2 = 1/3$ , confirming that A is correct. Choice B gives $1/\sqrt{3}$ , which is the ratio of the resultant amplitudes ( $A/A\sqrt{3}$ ), not intensities — a classic trap of forgetting to square. Choice C claims intensity changes in direct proportion to phase difference, but this is physically incorrect; the relationship is trigonometric, not linear. Choice D suggests a larger phase difference increases the resultant amplitude, but $\cos(60°) < \cos(30°)$ , so the opposite is true here — a larger phase difference reduced the resultant. A useful rule to memorize: amplitude ratios and intensity ratios are never the same. Whenever a question asks about intensity, automatically square your amplitude result before comparing.
A pulse carrying $5.0~J$ reaches a boundary between two strings. The reflected pulse travels back along the original string with an amplitude equal to $0.60$ of the incident amplitude. If the boundary absorbs no energy, how much energy is carried by the transmitted pulse?
Answer: $3.2~J$ , because the reflected energy fraction is the square of $0.60$ When a wave reaches a boundary between two media, its energy splits between a reflected pulse and a transmitted pulse. The critical concept here is that wave energy is proportional to the square of amplitude — not amplitude itself. So when amplitude changes by a factor, energy changes by that factor squared. Here's the reasoning for choice C: the reflected pulse has amplitude $0.60$ times the incident amplitude, so the reflected energy fraction is $(0.60)^2 = 0.36$ . This means the reflected pulse carries $0.36 \times 5.0~\text{J} = 1.8~\text{J}$ . Since the boundary absorbs no energy, conservation of energy gives the transmitted pulse: $5.0 - 1.8 = 3.2~\text{J}$ . Choice C is correct. Now let's clear up the traps. Choice A gets the reflected energy right (1.8 J) but mistakenly labels it as the transmitted energy — a straightforward mix-up between which pulse is which. Choice B assumes the transmitted amplitude fraction is simply $1 - 0.60 = 0.40$ , then squares it: $(0.40)^2 \times 5.0 = 0.8~\text{J}$ ... actually, B seems to use $0.40$ directly as an energy fraction giving $2.0~\text{J}$ , which confuses amplitude fractions with energy fractions entirely. Choice D assumes one-fifth of the energy is reflected (giving 1.0 J reflected and 4.0 J transmitted), but there's no basis for that — the reflected amplitude fraction of 0.60 is given, not an energy fraction of 0.20. Your key takeaway: always square the amplitude ratio to find the energy ratio. On wave problems, conflating amplitude with energy is the most common trap — expect it on every boundary/reflection question.

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