Apply Trig Waves to Sound & Light in Context

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Quizzes Created: 7202 | Total Attempts: 9,524,167
| Questions: 20 | Updated: Nov 10, 2025
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1) P(t) = 0.8 sin(440·2π t). Frequency?

Explanation

Argument = 2π·440 t ⇒ f = 440. Hence, 440 Hz.

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About This Quiz
Apply Trig Waves To Sound & Light In Context - Quiz

Explore how trigonometric wave equations apply to real-world examples like musical tones, LED lights, and oscillating signals. You will analyze given equations, identify features such as amplitude and period, and interpret how these relate to measurable wave behavior. This quiz reinforces the link between mathematical functions and the physical patterns... see moreseen in sound and light waves. see less

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2) I(t) = 5 + 2 cos(10π t). Midline?

Explanation

Form A cos(⋯)+D ⇒ midline y = D = 5. Hence, y = 5.

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3) Y(x,t) = 3 sin(2π(120 t − x/0.9)). Wave speed?

Explanation

Phase 2π(120 t − x/0.9) ⇒ v = 120 ÷ (1/0.9) = 108. Hence, 108 m/s.

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4) L(t) = 12 − 4 sin(π t − π/2). Phase shift (seconds)?

Explanation

sin(ωt − φ): shift = φ/ω = (π/2)/π = 0.5 s to the right. Hence, 0.5 s right.

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5) S(t) = A sin(2π f t) becomes amplitude doubled, frequency halved.

Explanation

New: 2A sin(2π (f/2) t) (≡ 2A sin(π f t)). Hence, D.

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6) λ = 500 nm, c = 3.0×10^8 m/s. Frequency?

Explanation

f = c/λ = 3e8 / 5e−7 = 6e14. Hence, 6.0×10^14 Hz.

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7) P(t) = 1.2 cos(600π t). Period?

Explanation

ω = 600π ⇒ f = 300 ⇒ T = 1/300 s. Hence, 1/300 s.

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8) Max y = 9, min y = 3, next max at t = 0.04 s. Which model?

Explanation

Amplitude = (9−3)/2 = 3; midline = 6; T = 0.04 ⇒ ω = 50π; max at t=0 ⇒ cosine. Hence, A.

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9) S1 = 0.5 sin(2π·440 t), s2 = 0.5 sin(2π·444 t). What is heard?

Explanation

Beat frequency = |444−440| = 4 Hz. Hence, beats at 4 Hz.

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10) I(t) = 10 + 5 sin(40π t + π/2). Which is true?

Explanation

Amplitude = 5, midline = 10. Hence, B.

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11) Y(t) = 2 sin(8π t − π/3). Frequency?

Explanation

2π f = 8π ⇒ f = 4. Hence, 4 Hz.

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12) S(t) = 3 sin(2π·220 t) + 3 sin(2π·220 t + π). Result?

Explanation

Phase difference π ⇒ destructive: A + (−A) = 0. Hence, silence.

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13) I(t) = 2 + 2 cos(2π t). At t = 0.25 s, I = ?

Explanation

cos(2π·0.25) = cos(π/2) = 0 ⇒ I = 2. Hence, 2.

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14) Amplitude 7, period 0.02 s; y(0) at midline and increasing.

Explanation

T=0.02 ⇒ ω=100π. Midline & increasing ⇒ sine starting upward. Hence, B.

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15) V = 2.0×10^8, f = 4.0×10^14. Wavelength?

Explanation

λ = v/f = 2e8 / 4e14 = 5e−7 m. Hence, 5.0×10−7 m.

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16) P(t) = 0.2 + 0.6 sin(100π t − π/2). Which statement is correct?

Explanation

Offset = 0.2, amplitude = 0.6. (Also f = 50 Hz.) Hence, A.

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17) Period 5 ms. Which ω fits?

Explanation

T=0.005 ⇒ f=200 ⇒ ω=2πf=400π rad/s. Hence, 400π.

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18) I(t) = I0 + k cos(2π t/T). If T halves (k,I0 same)…

Explanation

f = 1/T ⇒ halving T doubles f; amplitude k unchanged. Hence, B.

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19) Y(t) = 5 cos(6π t). First zero (smallest t > 0)?

Explanation

cos(6π t) = 0 ⇒ 6π t = π/2 ⇒ t = 1/12. Hence, 1/12 s.

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20) I(t) = 8 + 3 sin(2π·120 t). Which is true?

Explanation

Peak = 8 + 3 = 11. (Minimum = 5 also true.) Hence, A.

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P(t) = 0.8 sin(440·2π t). Frequency?
I(t) = 5 + 2 cos(10π t). Midline?
Y(x,t) = 3 sin(2π(120 t − x/0.9)). Wave speed?
L(t) = 12 − 4 sin(π t − π/2). Phase shift (seconds)?
S(t) = A sin(2π f t) becomes amplitude doubled, frequency halved.
λ = 500 nm, c = 3.0×10^8 m/s. Frequency?
P(t) = 1.2 cos(600π t). Period?
Max y = 9, min y = 3, next max at t = 0.04 s. Which model?
S1 = 0.5 sin(2π·440 t), s2 = 0.5 sin(2π·444 t). What is heard?
I(t) = 10 + 5 sin(40π t + π/2). Which is true?
Y(t) = 2 sin(8π t − π/3). Frequency?
S(t) = 3 sin(2π·220 t) + 3 sin(2π·220 t + π). Result?
I(t) = 2 + 2 cos(2π t). At t = 0.25 s, I = ?
Amplitude 7, period 0.02 s; y(0) at midline and increasing.
V = 2.0×10^8, f = 4.0×10^14. Wavelength?
P(t) = 0.2 + 0.6 sin(100π t − π/2). Which statement is correct?
Period 5 ms. Which ω fits?
I(t) = I0 + k cos(2π t/T). If T halves (k,I0 same)…
Y(t) = 5 cos(6π t). First zero (smallest t > 0)?
I(t) = 8 + 3 sin(2π·120 t). Which is true?
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