Chapter 7.2, Problem 69E

Interference Two identical tuning forks are struck, one a fraction of a second after the other. The sounds produced are modeled by f₁(t) = C sin ωt and f₂(t) = C sin(ωt + α). The two sound waves interfere to produce a single sound modeled by the sum of these functions

$f(t)=C\sin \omega t+C\sin (\omega t+\alpha )$

1. (a) Use the Addition Formula for Sine to show that f can be written in the form f(t) = A sin ωt + B cos ωt, where A and B are constants that depend on α.
2. (b) Suppose that C = 10 and α = π/3. Find constants k and ϕ so that f(t) = k sin(ωt + ϕ).

To determine

To show: The given function $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$ can be written in the form $f\left(t\right)=A\text{sin}\omega t+B\text{cos}\omega t$, where A and B are constants that depend on $\alpha$.

Explanation of Solution

Given:

The given function is $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$.

Formula used:

The addition formula of sine function is $\sin \left(a+b\right)=\sin a\cos b+\cos a\sin b$.

Proof:

Consider the given function $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$ (1)

Use the addition formula for sine function and expand $\text{sin}\left(\omega t+\alpha \right)$ as $\text{sin}\left(\omega t+\alpha \right)=\sin \omega t\cos \alpha +\cos \omega t\sin \alpha$.

Thus, equation (1) is rewritten as follows.

$\begin{array}{c}f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)\\ =C\text{sin}\omega t+C\left(\sin \omega t\cos \alpha +\cos \omega t\sin \alpha \right)\end{array}$

Rearrange to obtain that $f\left(t\right)=\left(C+C\cos \alpha \right)\text{sin}\omega t+\left(C\sin \alpha \right)\cos \omega t$.

Let $A=C+C\cos \alpha$ and $B\text{=}C\sin \alpha$

Thus, $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$ can be written in the form $f\left(t\right)=A\text{sin}\omega t+B\text{cos}\omega t$, where A and B are constants that depend on $\alpha$.

To determine

To find: The constants $k$ and $\varphi$ so that the given function satisfies $f\left(t\right)=k\sin \left(\omega t+\varphi \right)$, given $C\text{=10}$ and $\alpha =\frac{\pi }{3}$.

Answer to Problem 69E

For $C\text{=10}$ and $\alpha =\frac{\pi }{3}$, the given function $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$ satisfies the form of sine curve $f\left(t\right)=k\sin \left(\omega t+\varphi \right)$ for the values $k=10\sqrt{3}$ and $\varphi =\frac{\pi }{6}$.

Explanation of Solution

Given:

For the function $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$, $C\text{=10}$ and $\alpha =\frac{\pi }{3}$.

Formula used:

Pythagorean identity: ${\cos }^{2}x+{\sin }^{2}x=1$.

The addition formula of sine function is $\sin \left(a+b\right)=\sin a\cos b+\cos a\sin b$.

Calculation:

From the part (a) $f\left(t\right)=A\text{sin}\omega t+B\text{cos}\omega t$, where $A=C+C\cos \alpha$ and $B\text{=}C\sin \alpha$.

Substitute $C\text{=10}$ and $\alpha =\frac{\pi }{3}$ in $A=C+C\cos \alpha$ and $B\text{=}C\sin \alpha$.

$\begin{array}{c}A=C+C\cos \alpha \\ =10+10\cos \frac{\pi }{3}\\ =10+10\left(\frac{1}{2}\right)\\ =15\end{array}$

$\begin{array}{c}B\text{=}C\sin \alpha \\ =10\sin \frac{\pi }{3}\\ =10\left(\frac{\sqrt{3}}{2}\right)\\ =5\sqrt{3}\end{array}$

Thus, $A=15$ and $B=5\sqrt{3}$.

Substitute the value $A=15$ and $B=5\sqrt{3}$ in $f\left(t\right)=A\text{sin}\omega t+B\text{cos}\omega t$ and obtain the following.

$f\left(t\right)=15\text{sin}\omega t+5\sqrt{3}\cos \omega t$ (2)

For $f\left(t\right)=C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)$ to satisfy $f\left(t\right)=k\sin \left(\omega t+\varphi \right)$, set $C\text{sin}\omega t+C\text{sin}\left(\omega t+\alpha \right)=k\sin \left(\omega t+\varphi \right)$.

Use the addition formula for sine function and expand $k\sin \left(\omega t+\varphi \right)$ as $k\sin \left(\omega t+\varphi \right)=k\sin \omega t\cos \varphi +k\cos \omega t\sin \varphi$.

Thus, equation (2) is $15\text{sin}\omega t+5\sqrt{3}\cos \omega t=k\sin \omega t\cos \varphi +k\cos \omega t\sin \varphi$.

Compare both equations and obtain the following.

$\begin{array}{l}15\sin \omega t=k\sin \omega t\cos \varphi \\ k\cos \varphi =15\end{array}$

Similarly,

$\begin{array}{l}5\sqrt{3}\cos \omega t=k\cos \omega t\sin \varphi \\ k\sin \varphi =5\sqrt{3}\end{array}$

Use the Pythagorean identity and obtain ${\cos }^2\varphi +{\sin }^2\varphi =1$.

Multiply both side by $k^2$ and obtain the following.

$\begin{array}{l}{k}^2{\cos }^2\varphi +k^2{\sin }^2\varphi =k^2\\ {\left(k\cos \varphi \right)}^2+{\left(k\sin \varphi \right)}^2=k^2\end{array}$

Substitute the value $k\cos \varphi =15$ and $k\sin \varphi =5\sqrt{3}$ in ${\left(k\cos \varphi \right)}^2+{\left(k\sin \varphi \right)}^2=k^2$.

$\begin{array}{l}{\left(k\cos \varphi \right)}^2+{\left(k\sin \varphi \right)}^2=k^2\\ {\left(15\right)}^2+{\left(5\sqrt{3}\right)}^2=k^2\\ k^2=225+75=300\\ k=10\sqrt{3}\end{array}$

Substitute the value $k=10\sqrt{3}$ in $k\cos \varphi =15$ and $k\sin \varphi =5\sqrt{3}$,

Thus, $\cos \varphi =\frac{15}{10\sqrt{3}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}$ and $\sin \varphi =\frac{5\sqrt{3}}{10\sqrt{3}}=\frac{1}{2}$.

Since, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ and $\sin \frac{\pi }{6}=\frac{1}{2}$, the value of $\varphi$ is $\frac{\pi }{6}$

Hence, the given function $y=f\left(t\right)$ satisfies the form of sine curve $y=k\sin \left(t+\varphi \right)$ for the values $k=10\sqrt{3}$ and $\varphi =\frac{\pi }{6}$.