The zeroes of the equation are (1,4)
Step-by-step explanation:
Given equation is:
We have to put y=0 in the equation to find zeroes of the equation
Hence,
The zeroes of the equation are (1,4)
Keywords: Equation, zeroes of an equation
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Answer:
0.20 and 0.36
Step-by-step explanation:
y(t) = 2 sin (4π t) + 5 cos (4π t)
We wish to convert this to:
y = A sin(ωt + φ)
We know that ω = 4π. We also know the following:
5 = A sin φ
2 = A cos φ
Divide the first equation by the second equation:
5/2 = tan φ
φ = tan⁻¹(5/2)
Now, square the two equations and add them together.
5² + 2² = (A sin φ)² + (A cos φ)²
29 = A²
A = √29
The equation of the wave is therefore:
y = √29 sin(4π t + tan⁻¹(5/2))
The maximum negative position is -√29. And half of that is -½√29.
-½√29 = √29 sin(4π t + tan⁻¹(5/2))
-½ = sin(4π t + tan⁻¹(5/2))
7π/6 + 2kπ or 11π/6 + 2kπ = 4π t + tan⁻¹(5/2)
7 + 12k or 11 + 12k = 24t + 6 tan⁻¹(5/2) / π
t = (7 + 12k − 6 tan⁻¹(5/2) / π) / 24 or (11 + 12k − 6 tan⁻¹(5/2) / π) / 24
Trying different integer values of k, we find there are two possible values for t between 0 and 0.5, both when k = 0.
t = (7 − 6 tan⁻¹(5/2) / π) / 24 or (11 − 6 tan⁻¹(5/2) / π) / 24
t ≈ 0.20 or 0.36
