FAQ
TL;DR: v(t)=Vm·sin(ωt) models a sine voltage. Concrete stat: at 50 Hz, one cycle lasts 20 ms. Expert quote: "at 50 hertz it would be 20 millisec." [Elektroda, David Adams, post #21662739]
Why it matters: This FAQ helps beginners interpret sinewave equations, pick the right values, and avoid DC/RMS confusion in labs and exams, especially for AC mains and bench signals.
Quick Facts
- t is instantaneous time (seconds); v(t) is voltage at that instant, not over a whole period. [Elektroda, Steve Lawson, post #21662734]
- ω (omega) equals 2πf; write v(t)=Vm·sin(ωt) and use Vp‑p=2·Vm when needed. [Elektroda, David Adams, post #21662735]
- Typical mains timing: 60 Hz → 16.67 ms period; 50 Hz → 20 ms period. [Elektroda, David Adams, post #21662739]
- Average (rectified) value of a sine ≈ 0.637·Vm; RMS of a pure sine ≈ 0.707·Vm. [Elektroda, Steve Lawson, post #21662742]
- DC offset adds a constant: v(t)=Vdc+Vm·sin(ωt); zero DC offset means purely AC. [Elektroda, David Adams, post #21662735]
What do t and v mean in v=Vm·sin(ωt)?
t is instantaneous time in seconds. v or v(t) is the instantaneous voltage value at that time. v changes as t changes, following the sine. Write it explicitly as v(t) to emphasize that voltage depends on time. [Elektroda, David Adams, post #21662735]
Is t the time period T of the signal?
No. t is a running time variable. The time period T is one full cycle’s duration and equals 1/f. In contrast, t can be any moment you evaluate the waveform. Confusing t with T leads to wrong phase and amplitude results. [Elektroda, Steve Lawson, post #21662734]
What is ω and how does it relate to frequency f?
ω is angular frequency in radians per second. It equals 2πf, where f is in hertz. Substitute ω=2πf into v(t)=Vm·sin(ωt) whenever you know f instead of ω. [Elektroda, David Adams, post #21662735]
How do I compute the instantaneous voltage at a specific time?
Use v(t)=Vm·sin(2πft). Example: Vm=10 V, f=50 Hz, t=5 ms → v=10·sin(2π·50·0.005)=10·sin(π/2)=10 V. Pro tip: express t in seconds. [Elektroda, David Adams, post #21662735]
What’s the difference between DC offset and average value?
DC offset (DC component) is the algebraic average of the waveform over a cycle. A pure sine’s algebraic average is zero. The average value often cited in rectifier contexts means the average of the absolute value; for a sine it is ≈0.637·Vm. [Elektroda, Steve Lawson, post #21662742]
How is RMS different from DC offset or average value?
RMS measures heating-equivalent value. For a pure sine, Vrms≈0.707·Vm. DC offset is a constant added to the waveform. The rectified average (≈0.637·Vm) characterizes rectifier outputs, not the AC’s heating effect. [Elektroda, Steve Lawson, post #21662742]
What does peak-to-peak (Vp‑p) mean versus Vm?
Vm (often Vp) is the maximum excursion from zero. Peak‑to‑peak is the total swing, Vp‑p=2·Vm. If Vm=5 V, Vp‑p=10 V. Many oscilloscopes display Vp‑p by default, so convert as needed. [Elektroda, David Adams, post #21662735]
What happens over one 50 Hz cycle in time?
At 50 Hz, T=20 ms. Typical landmarks: 0 ms→0 V crossing, 5 ms→positive peak, 10 ms→0 V, 15 ms→negative peak, 20 ms→back to 0 V. Then it repeats. This timeline helps when checking phase on scopes. [Elektroda, David Adams, post #21662739]
Why can’t I see AC voltage changing with my eyes?
You can’t see electrical changes in wires. With a lamp driven by AC, flicker becomes noticeable below about 70 Hz, obvious near 30 Hz, and clear near 10 Hz. "What you see on a scope is the history of the event." [Elektroda, Steve Lawson, post #21662738]
How do I add a DC offset to a sine wave in the equation?
Include a constant term: v(t)=Vdc+Vm·sin(ωt). Vdc shifts the waveform up or down without changing its frequency or peak‑to‑peak. Zero offset means the positive and negative halves are symmetric. [Elektroda, David Adams, post #21662735]
Is V in v=Vm·sin(ωt) the DC offset?
No. In this equation, v (lowercase) is instantaneous voltage. If there is a DC offset, write v(t)=Vdc+Vm·sin(ωt). Using just v=Vm·sin(ωt) implies zero DC offset. [Elektroda, David Adams, post #21662735]
How can I tell if a waveform has a DC component?
Compute its algebraic average over one period. If the average is not zero, the waveform has a DC component (offset). A pure sine centered on zero has none. [Elektroda, Steve Lawson, post #21662742]
Quick how‑to: evaluate a sine at any time t
- Compute ω=2πf.
- Multiply angle θ=ωt (radians).
- Calculate v(t)=Vm·sin(θ); include +Vdc if present.
Use consistent units (seconds, hertz, volts). [Elektroda, David Adams, post #21662735]
Can an oscilloscope show these relationships clearly?
Yes. Timebase shows period T; cursors can measure 5 ms, 10 ms, etc., at 50 Hz. Vertical scaling shows Vm or Vp‑p so you can infer Vrms or average if the scope lacks math functions. [Elektroda, David Adams, post #21662739]
Do people often confuse t with T in practice? What’s the risk?
Yes. Treating t as the fixed period T misplaces phase and yields wrong instantaneous values. Always keep t as a variable and use T=1/f only for cycle length. [Elektroda, Steve Lawson, post #21662734]
