91Ó°ÊÓ

The frequency of the fundamental note of a vibrating string is given by the formula: \(f = \frac{1}{2L} \sqrt{\frac{T}{μ}}\) where f is the frequency, L is the length of the string, T is the tension, and μ is the linear mass density of the string.

According to the problem, the length of the wire is increased by half, i.e., the new length L' is: \(L' = 1.5 + \frac{1.5}{2} = 1.5 + 0.75 = 2.25 \mathrm{~m}\) We are asked to find the fundamental frequency when the length is increased to 2.25 meters under the same tension. Using the formula from Step 1, we get: \(f' = \frac{1}{2L'} \sqrt{\frac{T}{μ}}\) Since the tension and linear mass density remain the same, we can express the new frequency in terms of the initial frequency and lengths: \(f' = f \frac{L}{L'} = 120 \frac{1.5}{2.25} = 120 \cdot \frac{2}{3} = 80 \mathrm{~Hz}\) The fundamental frequency of the wire when the length is increased by half is 80 Hz.

The fundamental frequency when the length is increased by half under the same tension is 80 Hz.

We are asked to find by how much the length should be shortened so that the frequency increases three-fold. Let ∆L be the change in length (shortening), and let L'' be the new length: \(L'' = L - \Delta L\) The new frequency is three times the initial frequency: \(f'' = 3f = 3 \cdot 120 = 360 \mathrm{~Hz}\) Using the formula for the frequency, we get: \(360 = \frac{1}{2L''} \sqrt{\frac{T}{μ}}\) Since the tension and linear mass density remain the same: \(L'' = \frac{fL}{3f'}\) Plugging in the values: \(L'' = \frac{120 \cdot 1.5}{3 \cdot 360} = \frac{1.5}{9} = \frac{1}{6} \mathrm{~m}\) Now, we can find the change in length, ∆L: \(\Delta L = L - L'' = 1.5 - \frac{1}{6} = \frac{8}{6} = \frac{4}{3} \mathrm{~m}\)

The length of the wire should be shortened by \(\frac{4}{3}\) meters, or approximately 1.33 meters, to increase the frequency three-fold.