Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Answer:
Approximately [tex]9.62[/tex].
Explanation:
[tex]y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0][/tex].
[tex]y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)][/tex].
Notice that sine waves [tex]y_1[/tex] and [tex]y_2[/tex] share the same frequency and wavelength. The only distinction between these two waves is the [tex](-0.250)[/tex] in [tex]y_2\![/tex].
Therefore, the sum [tex](y_1 + y_2)[/tex] would still be a sine wave. The amplitude of [tex](y_1 + y_2)\![/tex] could be found without using calculus.
Consider the sum-of-angle identity for sine:
[tex]\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)[/tex].
Compare the expression [tex]\sin(a + b)[/tex] to [tex]y_2[/tex]. Let [tex]a = (4.35\, x - 1270)[/tex] and [tex]b = (-0.250)[/tex]. Apply the sum-of-angle identity of sine to rewrite [tex]y_2\![/tex].
[tex]\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Therefore, the sum [tex](y_1 + y_2)[/tex] would become:
[tex]\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Consider: would it be possible to find [tex]m[/tex] and [tex]c[/tex] that satisfy the following hypothetical equation?
[tex]\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Simplify this hypothetical equation:
[tex]\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}[/tex].
Apply the sum-of-angle identity of sine to rewrite the left-hand side:
[tex]\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}[/tex].
Compare this expression with the right-hand side. For this hypothetical equation to hold for all real [tex]x[/tex] and [tex]t[/tex], the following should be satisfied:
[tex]\displaystyle 1 + \cos(-0.250) = m\, \cos(c)[/tex], and
[tex]\displaystyle \sin(-0.250) = m\, \sin(c)[/tex].
Consider the Pythagorean identity. For any real number [tex]a[/tex]:
[tex]{\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2[/tex].
Make use of the Pythagorean identity to solve this system of equations for [tex]m[/tex]. Square both sides of both equations:
[tex]\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2[/tex].
[tex]\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2[/tex].
Take the sum of these two equations.
Left-hand side:
[tex]\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}[/tex].
Right-hand side:
[tex]\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}[/tex].
Therefore:
[tex]m^2 = 2 + 2\, \cos(-0.250)[/tex].
[tex]m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98[/tex].
Substitute [tex]m = \sqrt{2 + 2\, \cos(-0.250)}[/tex] back to the system to find [tex]c[/tex]. However, notice that the exact value of [tex]c\![/tex] isn't required for finding the amplitude of [tex](y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)[/tex].
(Side note: one possible value of [tex]c[/tex] is [tex]\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125[/tex] radians.)
As long as [tex]\! c[/tex] is a real number, the amplitude of [tex](y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)[/tex] would be equal to the absolute value of [tex](4.85\, m)[/tex].
Therefore, the amplitude of [tex](y_1 + y_2)[/tex] would be:
[tex]\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}[/tex].
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
