Exercises — Wave parameters — amplitude, wavelength, frequency, period, wave speed

What we look at: the snapshot graph has distance on the horizontal axis, so it shows how the wave repeats in space.

• Amplitude: the peak sits at $y=+0.3\text{m}$ above the middle line ($y=0$). Amplitude is measured from the middle, so $A=0.3\text{m}$.
• Wavelength: one crest is at $x=1\text{m}$, the next at $x=5\text{m}$. Crest-to-crest distance is $\lambda =5-1=4\text{m}$.

Answer: $A=0.3\text{m}$, $\lambda =4\text{m}$.

What we look at: this graph has time on the horizontal axis — one particle bobbing up and down. It shows repetition in time, so we read $T$ here, not $\lambda$.

• Amplitude: peak at $y=+0.3\text{m}$, so $A=0.3\text{m}$ (same idea as before — height above the middle line).
• Period: one full up-down-up cycle takes from $t=0$ to $t=2\text{s}$, so $T=2\text{s}$.
• Frequency: $f=1/T=1/2=0.5\text{Hz}$.

Answer: $A=0.3\text{m}$, $T=2\text{s}$, $f=0.5\text{Hz}$.

Why this tool: we know $v$ and $f$ and want $\lambda$; the wave relation $v=f\lambda$ links exactly these three. Rearrange for $\lambda$: $\lambda =\frac{v}{f}=\frac{340}{170}=2\text{m}.$ Answer: $\lambda =2\text{m}$.

Why this tool: we have $\lambda$ and $T$; speed is "one wavelength per period," i.e. $v=\lambda /T$. $v=\frac{\lambda }{T}=\frac{0.5}{0.02}=25\text{m/s}.$ Check via frequency: $f=1/T=1/0.02=50\text{Hz}$, then $v=f\lambda =50\times 0.5=25\text{m/s}$. ✓ Answer: $v=25\text{m/s}$.

• Frequency: $f=\frac{\text{cycles}}{\text{time}}=\frac{30}{6}=5\text{Hz}$.
• Period: $T=1/f=1/5=0.2\text{s}$.
• Wavelength: $\lambda =v/f=2/5=0.4\text{m}$. Answer: $f=5\text{Hz}$, $T=0.2\text{s}$, $\lambda =0.4\text{m}$.

Key insight: wave speed is set by the medium, not by how fast you shake. The string is unchanged, so $v$ is unchanged.

• Find the fixed speed: $v=f_1{\lambda }_1=40\times 0.6=24\text{m/s}$.
• New wavelength at the same $v$: ${\lambda }_2=v/f_2=24/120=0.2\text{m}$.
• Forecast check: frequency tripled ($40\to 120$), so wavelength should be one third ($0.6\to 0.2$). ✓ Answer: ${\lambda }_2=0.2\text{m}$.

Key insight: when a wave crosses a boundary, its frequency is fixed by the source (each wavefront that arrives must leave — you can't pile them up), but the speed changes with the medium. So $\lambda$ must change to keep $v=f\lambda$ true.

• ${\lambda }_1=v_1/f=300/200=1.5\text{m}$.
• ${\lambda }_2=v_2/f=450/200=2.25\text{m}$.
• Sense check: faster medium at the same frequency ⇒ longer wavelength. ✓ Answer: ${\lambda }_1=1.5\text{m}$, ${\lambda }_2=2.25\text{m}$.

• Amplitude: the $0.8\text{m}$ is peak-to-peak (lowest to highest $=2A$), so $A=0.8/2=0.4\text{m}$.
• Period: "lowest → highest → lowest" is one complete cycle, so $T=5\text{s}$.
• Frequency: $f=1/T=1/5=0.2\text{Hz}$.
• Wavelength: crest-to-crest gives $\lambda =15\text{m}$ directly.
• Speed: $v=f\lambda =0.2\times 15=3\text{m/s}$ (or $v=\lambda /T=15/5=3\text{m/s}$). ✓ Answer: $A=0.4\text{m}$, $T=5\text{s}$, $f=0.2\text{Hz}$, $v=3\text{m/s}$.

Match term by term with $y=A\sin (kx-\omega t)$:

• Amplitude: the number out front is $A=0.05\text{m}$.
• Wave number: $k=4\pi \text{rad/m}$. Since $k=2\pi /\lambda$, we get $\lambda =2\pi /k=2\pi /(4\pi )=0.5\text{m}$.
• Angular frequency: $\omega =100\pi \text{rad/s}$. Since $\omega =2\pi f$, we get $f=\omega /(2\pi )=100\pi /(2\pi )=50\text{Hz}$.
• Period: $T=1/f=1/50=0.02\text{s}$.
• Speed: $v=f\lambda =50\times 0.5=25\text{m/s}$ (equivalently $v=\omega /k=100\pi /4\pi =25\text{m/s}$). ✓ Answer: $A=0.05\text{m}$, $\lambda =0.5\text{m}$, $f=50\text{Hz}$, $T=0.02\text{s}$, $v=25\text{m/s}$.

Same tool ($v=f\lambda$) works for all waves; here $v=c$ is fixed.

• (a) Radio: ${\lambda }_a=c/f=(3.0\times 10^8)/(1.0\times 10^8)=3\text{m}$.
• (b) Green light: ${\lambda }_b=c/f=(3.0\times 10^8)/(6.0\times 10^{14})=5.0\times 10^{-7}\text{m}=500\text{nm}$.
• Comment: ${\lambda }_a/{\lambda }_b=3/(5\times 10^{-7})=6\times 10^6$ — radio waves are about six million times longer than green light, even though both zoom at the same speed $c$. Speed fixed by medium (vacuum) ⇒ frequency and wavelength trade off inversely, over an enormous range. Answer: ${\lambda }_a=3\text{m}$, ${\lambda }_b=5.0\times 10^{-7}\text{m}$.

• (a) $\lambda =v/f$. As $f\to 0^{+}$, the denominator shrinks toward zero, so $\lambda =v/f\to \infty$. The crests get pushed infinitely far apart.
• (b) $f=0$ means zero oscillations per second — the source never repeats. There is no travelling wiggle at all; it's a static (unchanging) disturbance, not a wave. So $f=0$ is a degenerate, non-wave limit — the formula still reports an infinite wavelength, which is the maths politely saying "no repetition in space either."
• (c) $A=0$ means the maximum displacement from the middle is zero: every particle stays put. There is no wave, whatever $f$ or $\lambda$ you write — amplitude is the "is anything actually moving?" parameter. A wave needs $A>0$. Answer: (a) $\lambda \to \infty$; (b) a static, non-oscillating disturbance (no true wave); (c) a flat, motionless medium (no wave).

• Speed check: $36\text{m}$ in $3\text{s}$ gives $v=36/3=12\text{m/s}$ ✓ (matches given).
• Wavelength from counting: $9$ wavelengths fit in the $36\text{m}$ travelled, so $\lambda =36/9=4\text{m}$.
• Frequency: $9$ crests pass the post in $3\text{s}$, so $f=9/3=3\text{Hz}$.
• Period: $T=1/f=1/3\approx 0.333\text{s}$.
• Independent confirm: $f\lambda =3\times 4=12\text{m/s}=v$. ✓ Two separate counts (distance-per-time and crests-per-time) agree — this is the wave relation in action. Answer: $\lambda =4\text{m}$, $f=3\text{Hz}$, $T=\frac{1}{3}\text{s}\approx 0.333\text{s}$.