Question

Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t − 0.250)] where x, y1, and y2 are in meters and t is in seconds. (a) What is the amplitude of the resultant wave function y1 + y2?

Answer 1

Approximately
`9.62`.

Step-by-step explanation:

`y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0]`.

`y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)]`.

Notice that sine waves
`y_1` and
`y_2` share the same frequency and wavelength. The only distinction between these two waves is the
`(-0.250)` in
`y_2\!`.

Therefore, the sum
`(y_1 + y_2)` would still be a sine wave. The amplitude of
`(y_1 + y_2)\!` could be found without using calculus.

Consider the sum-of-angle identity for sine:

`\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)`.

Compare the expression
`\sin(a + b)` to
`y_2`. Let
`a = (4.35\, x - 1270)` and
`b = (-0.250)`. Apply the sum-of-angle identity of sine to rewrite
`y_2\!`.

`\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}`.

Therefore, the sum
`(y_1 + y_2)` would become:

`\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}`.

Consider: would it be possible to find
`m` and
`c` that satisfy the following hypothetical equation?

`\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}`.

Simplify this hypothetical equation:

`\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}`.

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

`\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}`.

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real
`x` and
`t`, the following should be satisfied:

`\displaystyle 1 + \cos(-0.250) = m\, \cos(c)`, and

`\displaystyle \sin(-0.250) = m\, \sin(c)`.

Consider the Pythagorean identity. For any real number
`a`:

`{\left(\sin(a)\right)}^(2) + {\left(\cos(a)\right)}^(2) = 1^2`.

Make use of the Pythagorean identity to solve this system of equations for
`m`. Square both sides of both equations:

`\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2`.

`\displaystyle {\left(\sin(-0.250)\right)}^(2) = m^2\, {\left(\sin(c)\right)}^2`.

Take the sum of these two equations.

Left-hand side:

`\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}`.

Right-hand side:

`\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}`.

Therefore:

`m^2 = 2 + 2\, \cos(-0.250)`.

`m = √(2 + 2\, \cos(-0.250)) \approx 1.98`.

Substitute
`m = √(2 + 2\, \cos(-0.250))` back to the system to find
`c`. However, notice that the exact value of
`c\!` isn't required for finding the amplitude of
`(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)`.

(Side note: one possible value of
`c` is
`\displaystyle \arccos\left((1 + \cos(0.250))/(√(2 * (1 + \cos(0.250))))\right) \approx 0.125` radians.)

As long as
`\! c` is a real number, the amplitude of
`(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)` would be equal to the absolute value of
`(4.85\, m)`.

Therefore, the amplitude of
`(y_1 + y_2)` would be:

`\begin{aligned}|4.85\, m| &= 4.85 * √(2 + 2\, \cos(-0.250)) \\&\approx 9.62 \end{aligned}`.