Exercises — Beats — derivation, applications

(a) WHAT we do: subtract the two frequencies and drop the sign. $f_{\text{beat}}=|250-254|=4\text{ Hz}$ WHY: a beat is one full loud→soft→loud cycle, and those cycles happen at exactly the rate the two waves drift apart and re-line-up — which is the frequency difference. (b) The pitch is the average: $f_{\text{avg}}=\frac{250+254}{2}=252\text{ Hz}$ Answer: 4 beats/s, throbbing a 252 Hz note.

(a) FALSE. Same frequency $\Rightarrow f_{\text{beat}}=|f-f|=0$. Zero beats means the loudness never swells and fades — a perfectly steady note. (This is exactly the "tuning is finished" condition.) (b) FALSE. At $60$ Hz the swelling happens 60 times a second — far too fast for the ear to follow; you just hear two rough separate tones. At $3$ Hz you get a slow, clearly countable waah…waah…waah. Beats are audible only when $|f_1-f_2|$ is small (roughly under $10$ Hz).

(a) $f_{\text{beat}}=|440-436|=4$ Hz. (b) WHAT/WHY: one silence occurs once per beat cycle, so the gap between silences is the beat period $T_{\text{beat}}=1/f_{\text{beat}}$ — the reciprocal, because "4 events per second" means "one event every $1/4$ second". $T_{\text{beat}}=\frac{1}{4}=0.25\text{ s}$ Answer: 4 Hz; a silence every $0.25$ s.

Step 1 — beats per second: $f_{\text{beat}}=\frac{10\text{ beats}}{2\text{ s}}=5$ Hz. Step 2 — undo the absolute value. The formula $f_{\text{beat}}=|f_{\text{fork}}-500|$ hides a sign: $|x|=5$ has two solutions, $x=+5$ or $x=-5$. $f_{\text{fork}}=500+5=505\text{ Hz}\text{or}{f}_{\text{fork}}=500-5=495\text{ Hz}$ Answer: $495$ Hz or $505$ Hz. (We cannot yet tell which — that ambiguity is the whole point of L3.)

Step 1 — two candidates. $|f_X-320|=4\Rightarrow f_X=316$ or $324$ Hz. Step 2 — what wax does. Wax adds mass $m$. For a fork behaving like Simple Harmonic Motion, $f=\frac{1}{2\pi }\sqrt{k/m}$, so more mass ⇒ lower $f$. After waxing, $f_X$ drops. Step 3 — test each candidate against the observation "beats went 4 → 6" (further from 320):

• If $f_X=324$: waxing lowers it toward 320 → gap shrinks → beats would decrease. ✗ contradicts observation.
• If $f_X=316$: waxing lowers it away from 320 → gap grows → beats increase. ✓ matches "rose to 6". Answer: $f_X=316$ Hz. (Sanity: after waxing it sits at $320-6=314$ Hz, and $314<316$ — yes, it went down, consistent with added mass.)

Step 1: $|f_Y-384|=3\Rightarrow f_Y=381$ or $387$ Hz. Step 2 — what filing does. Filing removes mass, so $m$ decreases and $f=\frac{1}{2\pi }\sqrt{k/m}$ increases. After filing, $f_Y$ goes up. Step 3 — test each candidate against "beats went 3 → 1" (closer to 384):

• If $f_Y=381$: filing raises it toward 384 → gap shrinks → beats decrease. ✓
• If $f_Y=387$: filing raises it away from 384 → gap grows → beats increase. ✗ Answer: $f_Y=381$ Hz. The moral: wax lowers, filing raises; whichever way the beat count moves tells you which side of the standard the unknown started on. That single observation resolves the $\pm$ sign.

(a) Treat "sent" and "received" as the two superposing signals: $f_{\text{beat}}=|2000-2012|=12\text{ Hz}$ (b) By the Doppler Effect, an approaching reflector compresses the wave crests it bounces back, so each returning crest arrives sooner → higher frequency. (A receding target would give a lower returned frequency and the same-sized beat only if the shift magnitude were equal.) The $12$ Hz beat is the machine's direct, countable read-out of the submarine's speed.

(a) With $y_1=a\cos (2\pi \cdot 500t)$, $y_2=a\cos (2\pi \cdot 508t)$, apply $\cos C+\cos D=2\cos \frac{C-D}{2}\cos \frac{C+D}{2}$ (this is the same Superposition Principle algebra used for Standing Waves, just with time instead of space): $y(t)=2a\underset{\text{slow envelope}}{\underbrace{\cos (2\pi \cdot 4t)}}\underset{\text{fast heard tone}}{\underbrace{\cos (2\pi \cdot 504t)}}$ Here $\frac{508-500}{2}=4$ and $\frac{508+500}{2}=504$. (b) Envelope frequency $=4$ Hz. But loudness depends on $|A(t)|=|2a\cos (2\pi \cdot 4t)|$, and $|\cos |$ hits a peak twice per envelope cycle (once for the hump above zero, once for the dip below). So: $f_{\text{beat}}=2\times 4=8\text{ Hz}=|500-508|.✓$

Step 1 — use A as the anchor. A is highest, so B and C are both below A.

• A–B beat $=5$ and B is below A $\Rightarrow f_B=260-5=255$ Hz.
• A–C beat $=8$ and C is below A $\Rightarrow f_C=260-8=252$ Hz. Step 2 — check consistency with the B–C beat. $|f_B-f_C|=|255-252|=3\text{ Hz.}✓$ The three measurements are mutually consistent, confirming the picture. Answer: $f_B=255$ Hz, $f_C=252$ Hz. The "what if $2$ Hz" twist: If A–C were $2$ Hz then $f_C=260-2=258$ Hz, giving $|f_B-f_C|=|255-258|=3$ Hz — still consistent with the $3$-beat measurement! So the B–C beat alone cannot distinguish these worlds; the extra information "A is highest" plus the A–C beat is what pins the frequencies down. This is why real tuning uses a known reference: without an anchor the beat differences form a web with multiple sign-solutions.

Step 1 — Sound them together, count the beats. Say you hear $n$ beats/s. Then $|f_{\text{unknown}}-256|=n$, so the unknown is $256+n$ or $256-n$. Why: the absolute value gives two candidates; this measures the size of the gap. Step 2 — Wax the unknown fork, sound again, note whether beats rise or fall. Wax adds mass $m$, and $f=\frac{1}{2\pi }\sqrt{k/m}$ falls with more mass. So after waxing $f_{\text{unknown}}$ decreases. Step 3 — Decide the side:

• If beats decrease, the fork moved toward 256, so it was above 256 → $f_{\text{unknown}}=256+n$.
• If beats increase, the fork moved away from 256, so it was below 256 → $f_{\text{unknown}}=256-n$. Step 4 — Worked instance. Suppose Step 1 gives $n=6$ beats/s and Step 2 shows beats increasing to $\approx 8$/s. Then the fork is below 256: $f_{\text{unknown}}=256-6=250\text{ Hz}.$ Why this works: the wax converts the invisible $\pm$ sign into a visible change in a countable rhythm. This is the same comparator idea behind heterodyne radio and Amplitude Modulation — read a tiny frequency difference as a slow, measurable beat.

Beat frequency is the difference $|f_1-f_2|$; the pitch you hear is the average; a wax/file test resolves the sign the absolute value hides.