Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Eddie is experimenting with his sister's violin. Allow the "A" string of his sister's violin have an ultimate tensile strength $\sigma_{1}$. He tunes a string up to its highest possible frequency $f_{1}$ before it breaks. He then builds an exact copy of the violin, where all lengths have been increased by a factor of $\sqrt{2}$ and tunes the same string again to its highest possible frequency $f_{2}$. What is $f_{2} / f_{1}$ ? The density of the string does not change.

Note: The ultimate tensile strength is maximum amount of stress an object can endure without breaking. Stress is defined as $\frac{F}{A}$, or force per unit area.

Solution:
We note from a simple dimensional analysis that the angular frequency of the string $\omega$ will consist


of the tension $T$, the length of the string $L$ and the mass of the string $m$.

$$
\begin{aligned}
T & =\left[M L T^{-2}\right] \\
L & =[L] \\
m & =[M] \\
\omega & =\left[T^{-1}\right]
\end{aligned}
$$

Therefore, by rearranging, we find that

$$
\begin{aligned}
\omega & =T^{\alpha} L^{\beta} m^{\gamma} \\
{\left[T^{-1}\right] } & =\left[M L T^{-2}\right]^{\alpha}[L]^{\beta}[M]^{\gamma}
\end{aligned}
$$

Distributing the exponents, and rearranging gives us

$$
T^{-1}=M^{\alpha+\gamma} L^{\alpha+\beta} T^{-2 \alpha}
$$

We now have three equations

$$
\begin{aligned}
\alpha+\gamma & =0 \\
\alpha+\beta & =0 \\
-2 \alpha & =-1
\end{aligned}
$$

From here, we find that $\alpha=1 / 2$. Substituting this into the first equation gives us

$$
1 / 2+\gamma=0 \Longrightarrow \gamma=-1 / 2
$$

then substituting $\alpha$ into the second equation gives us

$$
1 / 2+\beta=0 \Longrightarrow \beta=-1 / 2
$$

We now find that the angular frequency is given by

$$
\omega=A \sqrt{\frac{T}{L m}}
$$

where $A$ is an arbritary constant. Noting that $\omega=2 \pi f$, we find that

$$
f=\frac{A}{2 \pi} \sqrt{\frac{T}{L m}}
$$

From this analysis, we can then see that $f_{2} / f_{1}=\sqrt{2} / 2 \approx 0.707$.

Final answer: $\frac{\sqrt{2}}{2}$