Question

Determine in what way and how many times will the fundamental tone frequency of a stretched wire change if its length is shortened by 35% and the tension increased by 70%.

Answer 1

The fundamental tone frequency will increase by approximately 2.006 times.

Answer 2

The fundamental frequency of a stretched wire is proportional to $\frac{1}{L}\sqrt{T}$. Given initial frequency $f_1 = \frac{1}{2L_1}\sqrt{\frac{T_1}{\mu}}$. New length $L_2 = 0.65 L_1$ and new tension $T_2 = 1.70 T_1$. The new frequency $f_2 = \frac{1}{2L_2}\sqrt{\frac{T_2}{\mu}} = \frac{1}{2(0.65 L_1)}\sqrt{\frac{1.70 T_1}{\mu}} = \left(\frac{\sqrt{1.70}}{0.65}\right) f_1$. Calculating the factor $\frac{\sqrt{1.70}}{0.65} \approx \frac{1.3038}{0.65} \approx 2.006$. Thus, the frequency increases by approximately 2.006 times.