Question

a person strikes one - end of a thin rod with a hammer. the speed of sound in the rod is 15 times the speed of sound in air. another person, at the other end with the ear close to the rod, hears the sound of the blow twice with a \\(\delta t = 0.12\\ s\\) interval between one sound comes through the rod and the other through air. if the sound speed in air \\(v = 331\\ m/s\\) at \\(0^{circ}c\\) and the temperature is \\(t = 30.0^{circ}c\\), what is the length of the rod?
\\(v=\lambda f =\frac{\delta x}{\delta t}\\)
\\(v_1=\frac{\delta x}{t_1}=\frac{l}{t_1}\\)
\\(v_2=\frac{\delta x}{t_2}=\frac{l}{t_2}\\)
\\(l =?\\)

Explanation:

Step1: Find the speed of sound in the rod

The speed of sound in air at $0^{\circ}C$ is $v_1 = 331\ m/s$. The speed of sound in air at $T = 30^{\circ}C$ can be calculated using the formula $v = v_0\sqrt{\frac{T + 273}{273}}$, where $v_0=331\ m/s$ and $T = 30^{\circ}C$. So $v_{air}=331\sqrt{\frac{30 + 273}{273}}\approx349\ m/s$. The speed of sound in the rod $v_{rod}=15\times v_{air}=15\times349 = 5235\ m/s$.

Step2: Set up time - difference equation

Let the length of the rod be $L$. The time taken for sound to travel through air $t_1=\frac{L}{v_{air}}$ and the time taken for sound to travel through the rod $t_2=\frac{L}{v_{rod}}$. We know that $t_1 - t_2=0.12\ s$. Substituting the expressions for $t_1$ and $t_2$ gives $\frac{L}{v_{air}}-\frac{L}{v_{rod}} = 0.12$.

Step3: Solve for the length of the rod

Factor out $L$: $L(\frac{1}{v_{air}}-\frac{1}{v_{rod}})=0.12$. Then $L=\frac{0.12}{\frac{1}{v_{air}}-\frac{1}{v_{rod}}}$. Substitute $v_{air}=349\ m/s$ and $v_{rod}=5235\ m/s$ into the formula:
\[

$$\begin{align*} L&=\frac{0.12}{\frac{1}{349}-\frac{1}{5235}}\\ &=\frac{0.12}{\frac{5235 - 349}{349\times5235}}\\ &=\frac{0.12\times349\times5235}{5235 - 349}\\ &=\frac{0.12\times349\times5235}{4886}\\ &\approx46.7\ m \end{align*}$$

\]

Answer:

$46.7\ m$