Sound Wave Function Quiz: Writing Basic Sound Wave Function

Substituting A = 1.2, F = 440, D = 0 gives s(t) = 1.2×sin(2π×440×t). Option B omits 2π, giving angular frequency 440 instead of 880π, corresponding to F ≈ 70 Hz. Option C uses π×440, giving F = 220 Hz. Option D uses 2πt/440, corresponding to F = 1/440 Hz. Only option A gives the correct angular frequency of 880π rad/s.

The angular frequency must be ω = 2πF = 2π×10 = 20π rad/s, giving s(t) = 6×sin(2π×10×t). Option A uses ω = 10, corresponding to F = 10/(2π) ≈ 1.59 Hz. Option B uses ω = 10π, corresponding to F = 5 Hz. Option C uses ω = 10 with a reflected amplitude. Only option D uses ω = 20π, correctly encoding F = 10 Hz with amplitude 6 and midline 0.

2.5 kHz = 2500 Hz. Substituting A = 0.8, F = 2500, D = 0 gives s(t) = 0.8×sin(2π×2500×t). Option B uses F = 2.5 Hz, failing to convert kHz to Hz. Option C uses F = 250 Hz, off by a factor of 10. Option D omits 2π, giving angular frequency 2500 instead of 5000π and F = 2500/(2π) ≈ 398 Hz. Only option A correctly converts 2.5 kHz to 2500 Hz and applies the standard form.

200 cycles per second means F = 200 Hz. Substituting A = 1.5, F = 200, D = 2 gives s(t) = 1.5×sin(2π×200×t) + 2. Option A omits 2π, giving F = 200/(2π) ≈ 31.8 Hz. Option B uses π×200 = 200π, giving F = 100 Hz. Option D uses 2πt/200, corresponding to F = 1/200 = 0.005 Hz. Only option C gives the correct angular frequency of 400π rad/s.

The answer is True. Amplitude is defined as abs(A), so abs(−A) = abs(A) — the amplitude is unchanged. The frequency depends only on F in the argument of sine, which is untouched by negating A. Negating A multiplies every output value by −1, flipping the wave symmetrically about the midline y = D. This is equivalent to a phase shift of π radians, which reflects the wave vertically without altering its shape, height, or repetition rate.

Substituting A = 4, F = 25, D = −3 gives s(t) = 4×sin(2π×25×t) − 3. Option A uses midline −1 instead of −3. Option C uses −4, reflecting the wave across the midline while keeping amplitude 4. Option D omits 2π, giving angular frequency 25 instead of 50π, corresponding to F = 25/(2π) ≈ 3.98 Hz, not 25 Hz.

Substituting A = 0.9, F = 880, D = 0.2 gives s(t) = 0.9×sin(2π×880×t) + 0.2. Option B uses D = −0.2, the wrong midline. Option C omits 2π, giving angular frequency 880 and F ≈ 140 Hz. Option D uses π×880 = 880π, giving F = 440 Hz instead of 880 Hz. Only option A correctly encodes all three parameters.

The answer is True. D is a vertical shift that moves every point on the graph up or down by D. Since sin oscillates symmetrically between −1 and +1, the scaled term A×sin oscillates symmetrically between −abs(A) and +abs(A). Adding D centers this oscillation on y = D. The output ranges from D − abs(A) to D + abs(A) symmetrically about the midline y = D.

Options A and B are both correct since 2π×3 = 6π, making them equivalent. Option C is incorrect because 2πt/3 gives angular frequency 2π/3, corresponding to F = 1/3 Hz, not 3 Hz. Option D is incorrect because the midline is 0 instead of the required D = 1. The question asks to identify errors, so C and D are the correct selections.

Replacing F with 2F gives angular frequency 2π×2F = 4πF, doubling the frequency. Replacing t with 2t gives argument 2π×F×2t = 4πFt, also doubling frequency. Keeping A and D the same while doubling F is a restatement of option A and remains valid. Option C replaces F with F/2, halving the frequency. Option D replaces t with t/2, halving the angular frequency and therefore halving the frequency.

From the standard form s(t) = A×sin(2πFt) + D: A = 2 confirms amplitude is 2. F = 50 Hz gives T = 1/F = 1/50 s. D = 3 is the midline. Maximum = D + A = 3 + 2 = 5. Option D is false — the coefficient 2π×50 identifies F = 50 Hz directly, not 100 Hz. The angular frequency ω = 2πF = 100π rad/s, but frequency in Hz is 50, not 100.

Matching to s(t) = A×sin(2πFt) + D gives A = 3, F = 75 Hz, D = 4. Option B incorrectly swaps A and F. Option C gives D = 0, ignoring the vertical shift of +4. Option D incorrectly reads F = 150, which would require the coefficient of t to be 2π×150 = 300π, but the coefficient is 2π×75 = 150π. Only option A correctly identifies all three parameters.

A = 0.3 confirms amplitude is 0.3. F = 1000 Hz gives T = 1/1000 = 0.001 s. D = −0.1 is the midline. Maximum = D + A = −0.1 + 0.3 = 0.2. All four statements correctly identify the parameters directly from the standard form of the equation.

The answer is True. The sine function always produces values between −1 and 1 inclusive. Multiplying by A scales this range to between −abs(A) and +abs(A). Adding D shifts the entire range upward by D, giving a final range from D − abs(A) to D + abs(A). This holds for all values of A and D without exception, confirming the amplitude and midline fully determine the wave's maximum and minimum values.

F = 1/T = 1/0.01 = 100 Hz. Substituting A = 2.5, F = 100, D = 1 gives s(t) = 2.5×sin(2π×100×t) + 1. Option B uses F = 10 Hz, giving T = 0.1 s, not 0.01 s. Option C uses F = 0.01 Hz, giving T = 100 s. Option D omits 2π, giving angular frequency 100 instead of 200π.

The angular frequency must be 2πF = 2π×60 = 120π rad/s, giving s(t) = 0.5×sin(2π×60×t) − 1. Option A uses 2πt/60, giving angular frequency 2π/60 which corresponds to F = 1/60 Hz, not 60 Hz. Option C uses π×60 = 60π, which gives F = 30 Hz. Option D omits 2π entirely, giving F = 60/(2π) ≈ 9.55 Hz.

The answer is True. Hz means cycles per second, so F has units of 1/second. For the product F×t inside the sine to be dimensionless, t must be in seconds. This makes F×t dimensionless, and multiplying by 2π converts cycles to radians, which is the required input for the sine function. Using any other unit for t would require a conversion factor to maintain dimensional consistency.

Substituting A = 3, F = 5, D = −2 into s(t) = A×sin(2πFt) + D gives s(t) = 3×sin(10πt) − 2. Option B uses D = +2, the wrong midline. Option C uses πF instead of 2πF, halving the angular frequency and doubling the period. Option D uses 2πt/5, giving angular frequency 2π/5 which corresponds to F = 0.2 Hz, not 5 Hz.

The answer is True. The argument of sine completes one full cycle when 2πFt increases by 2π. Setting 2πFT = 2π and solving gives T = 1/F. For example, F = 50 Hz gives T = 1/50 = 0.02 seconds, meaning the wave completes 50 full cycles every second. This relationship between period and frequency is fundamental to all sinusoidal wave models.

Options A and B are equivalent since 2π×4 = 8π, both giving amplitude 5 and midline −2. Option C uses −5 inside sine, but amplitude = abs(−5) = 5 and midline remains −2, so it is a valid reflected model. Option D uses 2πt/4, giving angular frequency π/2 which corresponds to F = 0.25 Hz, not 4 Hz.