Relate Wavelength Frequency Wave Speed - Grade 11 Physics

A wave travels 3.0 m in 0.50 s and has wavelength 1.5 m. What is its frequency?
Answer: 4.0 Hz First find the wave speed: 3.0 m / 0.50 s = 6.0 m/s. Then divide speed by wavelength: 6.0 / 1.5 = 4.0, so frequency is 4.0 Hz. A tempting wrong answer is 2.0 Hz, from dividing distance by wavelength directly, which ignores the 0.50 s travel time.
A device produces one pulse on a rope every $0.20\ \mathrm{s}$ . At one instant, six successive pulses are traveling along the rope, and the distance from the first pulse to the sixth pulse is $7.5\ \mathrm{m}$ . What is the pulse speed?
Answer: $7.5\ \mathrm{m/s}$ , using five time intervals between the pulses When a device emits pulses at regular intervals, the pulses spread out along the rope like evenly spaced beads on a string. The key insight is counting intervals, not pulses. Six pulses create only five gaps between them — think of fence posts and fences: six posts make five sections. The time between consecutive pulses is $T = 0.20\ \mathrm{s}$ , so the five intervals span a total time of $5 \times 0.20 = 1.0\ \mathrm{s}$ . The distance from pulse 1 to pulse 6 is $7.5\ \mathrm{m}$ . Applying the wave speed formula: $v = \frac{d}{t} = \frac{7.5\ \mathrm{m}}{1.0\ \mathrm{s}} = 7.5\ \mathrm{m/s}$ This confirms A is correct. Choice B makes the classic "fence post error" — counting six intervals instead of five. Six pulses means six time periods only if you're measuring from the moment the first pulse was emitted, which isn't the same as the gap between pulse 1 and pulse 6. Choice C arbitrarily halves the distance with no physical justification — there's no reason to use half the separation here. Choice D divides the full $7.5\ \mathrm{m}$ by just one period ( $0.20\ \mathrm{s}$ ), which would only be correct if the entire distance were covered in a single emission cycle, which it isn't. The study tip here: whenever you count objects and need the number of gaps between them, always subtract one. This "fence post" trap appears frequently in wave and pulse problems — train yourself to pause and ask, "Am I counting intervals or objects?"
For a periodic water wave, the distance from the second crest to the seventh crest is $15.0\ \mathrm{m}$ . At a fixed point, the time from the arrival of the first crest to the arrival of the twentieth crest is $9.5\ \mathrm{s}$ . What is the wave speed?
Answer: Approximately $6.00\ \mathrm{m/s}$ When solving wave problems like this, your two key formulas are $v = f\lambda$ and $v = \lambda/T$ . The trick is carefully counting the gaps between crests — not the crests themselves. From the second crest to the seventh crest, there are 5 gaps (2→3, 3→4, 4→5, 5→6, 6→7), so the wavelength is $\lambda = 15.0\ \mathrm{m} \div 5 = 3.0\ \mathrm{m}$ . For the period: from the first crest to the twentieth crest, there are 19 gaps, so the wave completes 19 full cycles in 9.5 s. The period is $T = 9.5\ \mathrm{s} \div 19 = 0.5\ \mathrm{s}$ , giving a frequency of $f = 1/T = 2.0\ \mathrm{Hz}$ . Wave speed: $v = f\lambda = 2.0\ \mathrm{Hz} \times 3.0\ \mathrm{m} = 6.0\ \mathrm{m/s}$ , confirming answer B. Choice A (5.00 m/s) likely results from dividing 15.0 m by 3 crests (counting the endpoints as gaps) for wavelength, then using the correct period — a classic off-by-one error. Choice C (6.32 m/s) comes from incorrectly counting 20 time intervals instead of 19, giving $T = 0.475\ \mathrm{s}$ and a slightly inflated speed. Choice D (5.26 m/s) combines both errors simultaneously — wrong gap count for both wavelength and period. The universal tip here: always count intervals, not endpoints. Whether counting crests or time gaps, subtract 1 from the number of reference points. This off-by-one trap appears frequently in wave problems on physics exams.
Light in glass has f = $5.0 \times 10^{14}$ Hz and refractive index n = 1.5. What is its wavelength in glass?
Answer: $4.0 \times 10^{-7}$ m Frequency stays the same as light enters glass, but speed drops to v/n. So wavelength is v/f = (3.0 x $10^8$ / 1.5) / (5.0 x $10^14$ ) = 4.0 x $10^{-7}$ m. The tempting 6.0 x $10^{-7}$ m is the wavelength in vacuum from c/f, because it ignores the slowing effect of the glass.
A wave produced by a constant-frequency source has a wavelength of $0.40\ \mathrm{m}$ and a frequency of $12\ \mathrm{Hz}$ in medium 1. It then enters medium 2, where its speed is $75\%$ of its speed in medium 1. What is its wavelength in medium 2?
Answer: $0.30\ \mathrm{m}$ , because wavelength changes with wave speed Whenever a wave crosses from one medium into another, you need to remember which properties change and which stay the same. The key principle: frequency is set by the source and never changes, but wave speed depends on the medium. Since speed and frequency are linked by $v = f\lambda$ , any change in speed must show up as a change in wavelength. Start by finding the speed in medium 1: $v_1 = f\lambda_1 = 12 \times 0.40 = 4.8\ \mathrm{m/s}$ . In medium 2, the speed drops to 75% of that: $v_2 = 0.75 \times 4.8 = 3.6\ \mathrm{m/s}$ . Since frequency stays at 12 Hz, the new wavelength is $\lambda_2 = \dfrac{v_2}{f} = \dfrac{3.6}{12} = 0.30\ \mathrm{m}$ . That confirms C is correct. A is wrong because it claims frequency decreases — frequency never changes when crossing media from a constant-frequency source. This is a classic misconception. B is partially right (frequency is indeed unchanged), but it incorrectly concludes that wavelength stays the same. Wavelength does change; only frequency is preserved. D gets the direction of change backwards — it says wavelength varies inversely with speed, which would mean slower speed gives a longer wavelength. In reality, wavelength and speed are directly proportional ( $\lambda = v/f$ ), so slower speed → shorter wavelength. A reliable strategy: memorize the mantra "frequency stays, speed changes, wavelength follows speed." When speed decreases by a factor, wavelength decreases by the same factor.
A wave hits a dock every 0.50 s and has wavelength 2.0 m. What is its speed?
Answer: 4.0 m/s The time between wave hits is the period, 0.50 s. Speed is wavelength divided by period, so 2.0 m divided by 0.50 s gives 4.0 m/s. The tempting wrong answer comes from multiplying wavelength by the period to get 1.0 m/s, but that confuses period with frequency.
If a wave's period doubles and its speed halves, what happens to its wavelength?
Answer: stays the same value Wavelength equals speed times period. If speed halves and period doubles, the changes cancel: half times double leaves the wavelength unchanged. A tempting mistake is to think halving speed alone halves wavelength, but the doubled period compensates for it.
A radar signal with a frequency of $2.40\ \mathrm{GHz}$ travels through a material at $2.00\times10^8\ \mathrm{m/s}$ . What is the wavelength of the signal in the material?
Answer: $8.33\times10^{-2}\ \mathrm{m}$ , obtained by correctly dividing the given speed by the given frequency When a wave travels through a medium, the relationship between its speed $v$ , frequency $f$ , and wavelength $\lambda$ is given by the fundamental wave equation: $v = f\lambda$ , which rearranges to $\lambda = \frac{v}{f}$ . The key here is using the wave's speed in the material, not in vacuum. Converting the frequency first: $2.40\ \mathrm{GHz} = 2.40 \times 10^9\ \mathrm{Hz}$ . Then applying the formula: $\lambda = \frac{2.00 \times 10^8\ \mathrm{m/s}}{2.40 \times 10^9\ \mathrm{Hz}} = 8.33 \times 10^{-2}\ \mathrm{m}$ This confirms C is correct. A is wrong because it multiplies speed and frequency ( $2.00 \times 10^8 \times 2.40$ ) rather than dividing. If you ever mix up which variable is being solved for, you'll fall into this trap — always rearrange $v = f\lambda$ explicitly before plugging in numbers. B uses the speed of light in vacuum ( $3.00 \times 10^8\ \mathrm{m/s}$ ) instead of the given speed of $2.00 \times 10^8\ \mathrm{m/s}$ . The problem explicitly tells you the wave is slowing down in the material — you must use that value. D results from a unit conversion error, treating 2.40 GHz as $1.00 \times 10^{10}\ \mathrm{Hz}$ rather than $2.40 \times 10^9\ \mathrm{Hz}$ . Always double-check your SI prefix conversions: Giga (G) means $\times 10^9$ , not $\times 10^{10}$ . Your go-to strategy: write out $\lambda = \frac{v}{f}$ , convert units carefully (especially GHz → Hz), and always use the speed given in the problem, not the speed of light in vacuum.
A sound wave has frequency 2.0 kHz and wavelength 0.17 m. What is its speed?
Answer: 340 m/s Wave speed equals frequency times wavelength. Convert 2.0 kHz to 2000 Hz, then multiply by 0.17 m: 2000 times 0.17 = 340, so the speed is 340 m/s. A tempting error is using 2.0 instead of 2000 Hz, which would give 0.34 m/s, but kilohertz must be converted to hertz first.
A sound generator maintains the same frequency while operating first in liquid A and then in liquid B. In liquid A, the distance from the first crest to the sixteenth crest is $22.5\ \mathrm{m}$ , and the wave speed is $1500\ \mathrm{m/s}$ . In liquid B, the distance from the first crest to the eleventh crest is $12.0\ \mathrm{m}$ . What is the wave speed in liquid B?
Answer: $1200\ \mathrm{m/s}$ , based on the ratio of wavelengths When a wave travels through a medium, the relationship $v = f\lambda$ connects wave speed, frequency, and wavelength. Since the generator keeps the same frequency in both liquids, any difference in wave speed must come entirely from a difference in wavelength. This makes finding each wavelength your first priority. To find the wavelength in liquid A, notice that the distance from the first crest to the sixteenth crest spans 15 gaps (not 16 — you count the intervals between crests, not the crests themselves). So $\lambda_A = \frac{22.5}{15} = 1.5\ \mathrm{m}$ . You can verify: $f = \frac{v_A}{\lambda_A} = \frac{1500}{1.5} = 1000\ \mathrm{Hz}$ . In liquid B, the first to the eleventh crest spans 10 gaps, giving $\lambda_B = \frac{12.0}{10} = 1.2\ \mathrm{m}$ . Since frequency is unchanged at 1000 Hz: $v_B = f\lambda_B = 1000 \times 1.2 = \boxed{1200\ \mathrm{m/s}}$ That confirms D is correct. A is wrong because subtracting crest counts (16 − 11 = 5) has no physical meaning here. B divides the raw distances directly (22.5 and 12.0) without accounting for the different number of gaps, so it skips the wavelength step entirely. C inverts the wavelength ratio before multiplying, producing a speed larger than liquid A's — a sign something flipped incorrectly. The key trap on wave problems is the off-by-one error: n crests always enclose n − 1 wavelengths. Burn that rule in before test day.
A student watches water waves pass a stationary buoy. The time from the arrival of the first crest to the arrival of the ninth crest is $16.0\ \mathrm{s}$ . The distance between adjacent crests is $6.0\ \mathrm{m}$ . What is the wave speed?
Answer: Approximately $3.00\ \mathrm{m/s}$ When a question involves wave speed, your three key quantities are speed, frequency, and wavelength, connected by the formula $v = f\lambda$ , where frequency $f = 1/T$ and $T$ is the period. The subtle trap here is correctly identifying the period from the given time information. The problem says the first crest arrives, then the ninth crest arrives 16.0 s later. It's tempting to divide 16.0 s by 9, but think carefully: from the 1st crest to the 9th crest, 8 intervals pass (1→2, 2→3, … 8→9), not 9. So the period is $T = 16.0\ \mathrm{s} \div 8 = 2.0\ \mathrm{s}$ . The wavelength (distance between adjacent crests) is $\lambda = 6.0\ \mathrm{m}$ . Therefore: $v = \frac{\lambda}{T} = \frac{6.0\ \mathrm{m}}{2.0\ \mathrm{s}} = 3.00\ \mathrm{m/s}$ , confirming answer B. Choice A (≈ 2.67 m/s) comes from incorrectly using 9 intervals instead of 8, giving $T \approx 1.78\ \mathrm{s}$ ... wait — actually A results from dividing $6.0 \div (16/6)$ , a garbled setup. More precisely, it appears from using $T = 16/9 \approx 1.78\ \mathrm{s}$ , yielding $v \approx 3.38\ \mathrm{m/s}$ — which is actually choice C. So C is the classic off-by-one error (dividing by 9 instead of 8). Choice A likely arises from a different algebraic misstep, such as dividing the total time by the wavelength times some factor. Choice D (9.00 m/s) ignores the period entirely and simply multiplies $6.0 \times 1.5$ or misapplies numbers without a coherent formula. Study tip: Whenever you count intervals between events, remember the fence-post rule — n objects create n − 1 gaps. From crest 1 to crest 9, there are 8 periods, not 9.
A wave's frequency increases by $20\%$ while its speed decreases by $10\%$ . What is the resulting percentage change in its wavelength?
Answer: The wavelength decreases by exactly $25\%$ Whenever you see a wave problem involving frequency, speed, and wavelength changing simultaneously, your anchor is the fundamental wave equation: $v = f\lambda$ , which rearranges to $\lambda = \frac{v}{f}$ . When frequency increases by $20\%$ , the new frequency is $1.20f$ . When speed decreases by $10\%$ , the new speed is $0.90v$ . Plugging these into the wavelength formula gives the new wavelength: $\lambda_{new} = \frac{0.90v}{1.20f} = \frac{0.90}{1.20} \cdot \frac{v}{f} = 0.75\lambda$ Since the new wavelength is $0.75\lambda$ , it has decreased by exactly $25\%$ , confirming that D is correct. Now let's see where the wrong answers lead you astray. A claims an increase — this is the opposite direction entirely, likely from confusing the relationship and thinking higher frequency means longer wavelength. B suggests a $10\%$ decrease, which only accounts for the speed change while completely ignoring the frequency increase — a classic partial-calculation trap. C arrives at $30\%$ by simply adding the two percentage changes ( $20\% + 10\% = 30\%$ ), which is tempting but mathematically incorrect — you cannot add percentage changes in multiplicative relationships; you must multiply the scale factors directly. Your strategy here: whenever multiple quantities change by percentages, always convert to scale factors and multiply (e.g., $0.90 \div 1.20$ ), never add or subtract the percentages. This rule applies broadly across physics and chemistry problems involving proportional relationships.
A laser has a wavelength of $600\ \mathrm{nm}$ in air, where its speed is approximately $3.00\times10^8\ \mathrm{m/s}$ . It enters a transparent slab in which its speed is $2.00\times10^8\ \mathrm{m/s}$ . Approximately how many wavelengths fit across a slab thickness of $0.200\ \mathrm{m}$ ?
Answer: $5.00\times10^5$ wavelengths, using the wavelength in the slab When light travels from one medium into another, its frequency stays constant but its wavelength changes proportionally with speed. This is the key insight for any question about light entering a new medium. To find how many wavelengths fit across the slab, you need the wavelength inside the slab, not in air. Since frequency is constant, you can use the ratio of speeds to find the new wavelength: $\lambda_{\text{slab}} = \lambda_{\text{air}} \times \frac{v_{\text{slab}}}{v_{\text{air}}} = 600\ \text{nm} \times \frac{2.00\times10^8}{3.00\times10^8} = 400\ \text{nm}$ Now divide the slab thickness by this wavelength: $N = \frac{0.200\ \text{m}}{400\times10^{-9}\ \text{m}} = 5.00\times10^5$ That confirms B is correct — you must use the compressed wavelength inside the medium. A is wrong because it uses the air wavelength (600 nm) inside the slab. Since the light slows down, its wavelength shrinks, so more wavelengths fit — not fewer. Using the air wavelength underestimates the count. C inverts the speed ratio incorrectly, producing a wavelength larger than the one in air, which is physically impossible for a denser medium where light slows down. D assumes frequency doubles when entering the medium, which never happens. Frequency is determined by the source and is invariant across media — only wavelength and speed change together. A quick study tip: whenever light changes media, anchor yourself to the fact that frequency is the traveler that never changes. Speed and wavelength shift together in the same direction, so slower speed always means shorter wavelength.
After the conditions of a wave-producing system change, the wave frequency becomes twice its original value while its wavelength becomes $25\%$ smaller. How does the new wave speed compare with the original wave speed?
Answer: The new speed is $1.50$ times the original speed Whenever you see a question about wave speed changing, your anchor should be the fundamental wave equation: $v = f\lambda$ . Wave speed is simply the product of frequency and wavelength, so if either — or both — change, you multiply the new values together to find the new speed. Here, the frequency doubles: $f_{new} = 2f$ . The wavelength becomes 25% smaller, meaning it's reduced to 75% of its original value: $\lambda_{new} = 0.75\lambda$ . Plugging into the wave equation gives you the new speed: $v_{new} = f_{new} \times \lambda_{new} = (2f)(0.75\lambda) = 1.50 \cdot f\lambda = 1.50 \cdot v_{original}$ The new speed is 1.50 times the original, confirming answer C. Now for the traps. Answer A (0.50×) likely comes from focusing only on the wavelength decrease and mistakenly taking half of 1.0 rather than computing the full product — it ignores the frequency change entirely. Answer B (0.75×) is the result of using only the wavelength factor (0.75) and completely ignoring that frequency doubled — a classic one-variable mistake. Answer D (2.25×) appears if you square the frequency factor ( $2^2 = 4$ ... or perhaps $1.5^2 = 2.25$ ), suggesting some students try to apply kinetic energy formulas or incorrectly square the speed ratio — neither is appropriate here. Study tip: Always write out $v = f\lambda$ first, then substitute each changed quantity as a multiplier of the original. This prevents you from accidentally ignoring one of the variables, which is exactly what most wrong answers on this type of question exploit.
Wave speed increases by 20% while frequency decreases by 10%. What happens to wavelength?
Answer: Increases by 33% Wavelength equals wave speed divided by frequency. New wavelength is 1.20 v divided by 0.90 f, which is 1.333 times the original, so it increases by about 33%. A common trap is just combining 20% and 10% as a 10% increase, but you must divide the new speed by the new frequency, not add percentages.
A 100 MHz radio wave travels at $3.0\times10^8$ m/s. Its wavelength is:
Answer: $3.0\times10^0$ m Convert 100 MHz to 1.0 x $10^8$ Hz. Then wavelength = speed / frequency = (3.0 x $10^8$ m/s) / (1.0 x $10^8$ Hz) = 3.0 m, or 3.0 x $10^0$ m. The 3.3 x $10^{-1}$ m answer comes from dividing frequency by speed, which gives 1/wavelength, not wavelength.
Two waves travel through the same uniform rope under unchanged tension. Wave A has a frequency of $8.0\ \mathrm{Hz}$ and a wavelength of $0.75\ \mathrm{m}$ . Wave B has a frequency of $12.0\ \mathrm{Hz}$ . What is the wavelength of wave B?
Answer: $0.50\ \mathrm{m}$ , because both waves have the same speed Whenever you see a question about waves traveling through the same medium, the key concept to activate is wave speed. In any uniform medium under constant tension, all waves travel at the same speed — frequency and wavelength can both vary, but their product is always fixed. The fundamental wave equation is $v = f\lambda$ . For Wave A: $v = 8.0\ \mathrm{Hz} \times 0.75\ \mathrm{m} = 6.0\ \mathrm{m/s}$ . Since Wave B travels through the same rope under the same tension, it must also travel at $6.0\ \mathrm{m/s}$ . Solving for Wave B's wavelength: $\lambda_B = \dfrac{v}{f_B} = \dfrac{6.0}{12.0} = 0.50\ \mathrm{m}$ . This confirms A is correct. B is wrong because subtracting the frequencies ( $12.0 - 8.0 = 4.0\ \mathrm{Hz}$ ) has no physical meaning here — you can't find a wavelength by taking a frequency difference. That operation is simply not part of the wave equation. C gets the relationship backwards. Frequency and wavelength are inversely proportional when speed is constant — higher frequency means shorter wavelength, not longer. Wave B has a higher frequency than Wave A, so its wavelength must be smaller than $0.75\ \mathrm{m}$ . D makes a similar inverse-relationship error and arrives at an even larger (and more incorrect) wavelength of $1.50\ \mathrm{m}$ . A reliable study tip: whenever waves share the same medium, write $v = f\lambda$ first and immediately calculate the shared speed. That single step prevents nearly every distractor trap on questions like this.
A wave has period 0.50 s and wavelength 1.5 m. How far does a crest move in 4.0 s?
Answer: 12 m Wave speed is wavelength divided by period: 1.5 / 0.50 = 3.0 m/s. In 4.0 s, a crest moves speed times time: 3.0 * 4.0 = 12 m. The tempting 3.0 m answer is just the speed, not the distance traveled over the full time.
A radio wave has frequency 98.5 MHz. What is its wavelength in air?
Answer: 3.05 m Radio waves travel at 3 x $10^8$ m/s in air. Divide that by 98.5 million Hz: 300,000,000 / 98,500,000 = 3.05 m. The common mistake is 0.305 m, which comes from using 985 million Hz instead of 98.5 million; MHz means $10^6$ , so 98.5 MHz is 98.5 million, not 985 million.
A wave's speed becomes 2/3 of its original value while its frequency stays the same. What happens to the wavelength?
Answer: It becomes 2/3 as large Wave speed equals frequency times wavelength, and frequency isn't changing, so wavelength must change in the same proportion as speed. If speed becomes 2/3 of its original value, wavelength becomes 2/3 as large too. The tempting wrong answer is 3/2 as large, but that would mean speed increased or frequency changed, neither of which happens.

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