Question

What would be the apparent frequency of a sound wave represented by

_{$\mathit{p}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{60}\mathbf{n}\mathbf{\pi }\mathbf{t}}{\mathbf{100}{\left(n-3\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{?}$}

Short answer

The apparent frequency of a sound waveis 90Hz.

Step-by-step solution

Given Information.

The given equation is$p\left(t\right)=\sum _{n=1}^{\infty }\frac{\cos 60n\mathrm{\pi }t}{100{\left(n-3\right)}^2+1}?$

Step 2: Meaning of the apparent frequency of a sound wave.

When the source moves with velocity towards the stationary listener, the perceived frequency of sound heard by the listener is ten percent higher than the actual frequency of the note emitted by the source.

Find the apparent frequency of a sound wave.

The function$a_n=n$then the function has a sharp spike at n=3

withrole="math" localid="1664373395549" $a_3=1$, and the next largest terms are:

$$\begin{gathered} a_2=a_4 \\ =\frac{1}{101} \end{gathered}$$

Hence, they are negligible.

Thus, the apparent frequency is

$$\begin{gathered} 180\mathrm{\pi t}=\frac{2\mathrm{\pi t}}{\mathrm{T}} \\ \mathrm{T}=\frac{1}{90} \end{gathered}$$

Since $T=\frac{1}{f}$

Therefore, the apparent frequency of a sound waveis 90Hz .