Tuning Fork 440Hz Quiz: Tuning Fork 440Hz Period and Model

Substituting A = 0.6, F = 440, φ = π/2, D = −0.2 into s(t) = A×sin(2πFt + φ) + D gives s(t) = 0.6×sin(2π×440×t + π/2) − 0.2. Option B uses phase π instead of π/2. Option C has the wrong midline +0.2 instead of −0.2. Option D uses 2πt/440, giving F = 1/440 Hz instead of 440 Hz. Only option A correctly encodes all four parameters.

Option A is the direct standard form. Option B uses 880π = 2π×440, making it identical to option A. Option D uses −sin, which reflects the wave but keeps amplitude = abs(−1) = 1 and midline 0. Option C uses 2πt/440, giving angular frequency 2π/440 and F = 1/440 Hz, not 440 Hz. Option D adds +1 to the midline, shifting it from 0 to 1.

T = 1/440 = 0.002272727... s, which rounds to 0.002273 s at 6 decimal places. Option B gives 0.004545 s = 1/220, the period of a 220 Hz tone. Option C gives 0.001136 s ≈ 1/880, the period of an 880 Hz tone. Option D gives 0.002500 s = 1/400, corresponding to a 400 Hz tone. Only 0.002273 s correctly rounds T = 1/440 to 6 decimal places.

The answer is True. Hz is defined as cycles per second, meaning F has units of 1/second. Multiplying F by t in seconds gives F×t in cycles, which is dimensionless. Multiplying by 2π then converts cycles to radians, which are also dimensionless and required as input for the sine function. Using any other unit for t without conversion would make F×t carry units rather than being dimensionless, breaking dimensional consistency.

T = 1/F = 1/220 s ≈ 0.004545 s. Option A gives 1/440 s, the period of a 440 Hz tone, which is double the frequency and therefore half the period. Option C gives 1/110 s, the period of a 110 Hz tone. Option D gives 1/880 s, the period of an 880 Hz tone. Only 1/220 s correctly applies T = 1/F with F = 220 Hz.

Option A is the direct standard form with A = 2, F = 440, D = 0. Option B uses −2, but amplitude = abs(−2) = 2 and midline remains 0, making it a valid reflected model. Option C uses 880π = 2π×440, so it is equivalent to option A. Option D uses 2πt/440, giving angular frequency 2π/440 and F = 1/440 Hz — incorrect. 

Substituting A = 0.8, F = 440, D = 0 gives s(t) = 0.8×sin(2π×440×t). Option B omits 2π, giving angular frequency 440 instead of 880π and F ≈ 70 Hz. Option C uses 2πt/440, giving F = 1/440 Hz. Option D uses π×440 = 440π, giving F = 220 Hz instead of 440 Hz. Only option A produces the correct angular frequency of 880π rad/s.

The answer is True. One complete cycle requires the argument 2πFt to increase by exactly 2π. Setting 2πFT = 2π and solving for T gives T = 1/F. This relationship holds for all values of A, F, and D. Neither the amplitude A nor the midline D appears in the argument of sine, so they have no effect on the period. The period is determined solely by F.

Substituting A = 1.2, F = 440, D = 0.05 into s(t) = A×sin(2πFt) + D gives s(t) = 1.2×sin(2π×440×t) + 0.05. Option B uses D = −0.05, the wrong midline. Option C omits 2π, giving angular frequency 440 instead of 880π and F ≈ 70 Hz. Option D uses 2πt/440, giving F = 1/440 Hz instead of 440 Hz. Only option A correctly encodes all three parameters.

Option A is correct — T = 1/440 ≈ 0.002273 s. Option B is correct — ω = 2π×440 = 880π rad/s. Option C is incorrect — 1/220 is twice the true period of 1/440 s, making this a false statement. Option D is correct — replacing 440 with 880 gives F = 880 Hz, doubling the frequency. C is the incorrect statement.

The answer is True. Consecutive peaks of a sine wave are exactly one full cycle apart. Since the period T is defined as the time for one complete cycle, the time elapsed between any two adjacent peaks equals exactly T. This holds for any sinusoidal wave regardless of amplitude, midline, or phase shift, as none of those parameters affect the horizontal spacing between peaks.

0.44 kHz = 440 Hz. Substituting A = 1, F = 440, D = 0 gives s(t) = sin(2π×440×t). Option A uses F = 0.44 Hz, failing to convert kHz to Hz. Option C omits 2π, giving angular frequency 440 instead of 880π and F ≈ 70 Hz. Option D uses 2πt/0.44, giving angular frequency 2π/0.44 and F ≈ 2.27 Hz. Only option B correctly converts 0.44 kHz to 440 Hz.

T = 1/F. T_440 = 1/440 ≈ 0.00227 s and T_220 = 1/220 ≈ 0.00455 s. Since 1/440 < 1/220, the 440 Hz tone has the shorter period. A higher frequency means the wave completes more cycles per second, so each cycle takes less time. The 440 Hz tone oscillates twice as fast as the 220 Hz tone, giving it exactly half the period.

T = 1/440 = 0.00227272727... s, which rounds to 0.00227273 s to 6 significant figures. Option B gives 1/220 = 0.00454545 s, the period of a 220 Hz tone. Option C gives 1/880 = 0.00113636 s, the period of an 880 Hz tone. Option D gives 0.0025 s, corresponding to F = 400 Hz. Only option A correctly applies T = 1/440 and rounds to 6 significant figures.

Substituting A = 0.3, F = 440, D = 0.1 into s(t) = A×sin(2πFt) + D gives s(t) = 0.3×sin(2π×440×t) + 0.1. Option B uses D = −0.1, the wrong midline. Option C omits 2π, giving angular frequency 440 instead of 880π and F ≈ 70 Hz. Option D uses 2πt/440, giving angular frequency 2π/440 and F = 1/440 Hz instead of 440 Hz.

The answer is True. The frequency F and period T = 1/F depend exclusively on the coefficient of t inside the sine function, which is 2πF. The amplitude A is a vertical scaling factor and D is a vertical shift. Neither A nor D appears in the argument of sine, so changing either one has no effect on how fast the wave oscillates. The wave's horizontal repetition rate is entirely determined by F.

Replacing F with 2F changes the angular frequency to 2π×2F = 4πF, doubling frequency and halving T. Replacing t with 2t changes the argument to 2π×F×2t = 4πFt, also doubling frequency and halving T. Keeping A and D while doubling F is equivalent to option A and confirms the period halves. Option C replaces F with F/2, halving frequency and doubling T. Option D replaces t with t/2, halving angular frequency and doubling T.

The answer is True. T = 1/F. Replacing F with 2F gives T_new = 1/(2F) = T/2. The period and frequency are reciprocals of each other, so any multiplicative change in one produces the inverse change in the other. Doubling the frequency compresses the wave horizontally so that twice as many cycles fit in the same time span, which is exactly what halving the period means.

F = 440 Hz gives T = 1/440 s, confirming A. Angular frequency ω = 2πF = 2π×440 rad/s, confirming B. Amplitude A = 1.5 means maximum = D + A = 0 + 1.5 = 1.5, confirming C. D = 0 is the midline, confirming D. All four statements are correct.

T = 1/F = 1/440 s ≈ 0.002273 s. Option B gives T = 1/220 s, which is the period of a 220 Hz tone, half the frequency of 440 Hz. Option C gives T = 1/880 s, corresponding to an 880 Hz tone, double the frequency. Option D gives T = 2/440 = 1/220 s, the same incorrect value as option B. Only 1/440 s correctly applies T = 1/F with F = 440 Hz.