Math Snap

STEP 1

1. The frequency of a vibrating string is affected by its length, thickness, and tension.
2. The frequency $f$ is given for a 2.0-ft-long string with a thickness of $\frac{1}{32}$ inch.
3. We need to find the frequency for a 1.6-ft-long string with a thickness of $\frac{3}{32}$ inch under two different tension conditions.

STEP 2

1. Understand the relationship between frequency, length, thickness, and tension.
2. Calculate the frequency under equal tension.
3. Calculate the frequency under one-third tension.

STEP 3

The frequency $f$ of a string is given by the formula:
$$f \propto \frac{1}{L} \sqrt{\frac{T}{\mu}}$$ where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density. The linear mass density $\mu$ is proportional to the cross-sectional area, which depends on the thickness.

STEP 4

For equal tension:
- Original string: $L_1 = 2.0$ ft, thickness $t_1 = \frac{1}{32}$ inch
- New string: $L_2 = 1.6$ ft, thickness $t_2 = \frac{3}{32}$ inch
The frequency ratio is:
$$\frac{f_2}{f_1} = \frac{L_1}{L_2} \times \sqrt{\frac{\mu_1}{\mu_2}}$$ Since $\mu \propto \text{thickness}^2$:
$$\frac{\mu_1}{\mu_2} = \left(\frac{t_1}{t_2}\right)^2 = \left(\frac{1/32}{3/32}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$$ Thus:
$$\frac{f_2}{f_1} = \frac{2.0}{1.6} \times \sqrt{\frac{1}{9}} = \frac{5}{4} \times \frac{1}{3} = \frac{5}{12}$$ So, $f_2 = \frac{5}{12} f$.

For one-third tension:
The frequency ratio considering tension $T_2 = \frac{1}{3} T_1$:
$$\frac{f_2}{f_1} = \frac{L_1}{L_2} \times \sqrt{\frac{\mu_1}{\mu_2}} \times \sqrt{\frac{T_2}{T_1}} = \frac{5}{12} \times \sqrt{\frac{1}{3}}$$ $$= \frac{5}{12} \times \frac{1}{\sqrt{3}} = \frac{5}{12\sqrt{3}}$$ So, $f_2 = \frac{5}{12\sqrt{3}} f$.
The frequencies are:
(a) $\frac{5}{12} f$ under equal tension.
(b) $\frac{5}{12\sqrt{3}} f$ under one-third tension.