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Math  /  Algebra

Question(8) A 2.0-ft-long nylon string with a thickness of 1/321 / 32 inch produces a frequency ff when plucked. What frequency would a 1.6 - ft -long nylon string that has a thickness of 3/323 / 32 inch produce (a) if put under an equal tension? (b) if put under only one-third of the original tension?

Studdy Solution

STEP 1

1. The frequency of a vibrating string is affected by its length, thickness, and tension.
2. The frequency f f is given for a 2.0-ft-long string with a thickness of 132 \frac{1}{32} inch.
3. We need to find the frequency for a 1.6-ft-long string with a thickness of 332 \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 f of a string is given by the formula:
f∝1LTμf \propto \frac{1}{L} \sqrt{\frac{T}{\mu}}
where L L is the length, T T is the tension, and μ \mu is the linear mass density. The linear mass density μ \mu is proportional to the cross-sectional area, which depends on the thickness.

STEP 4

For equal tension:
- Original string: L1=2.0 L_1 = 2.0 ft, thickness t1=132 t_1 = \frac{1}{32} inch - New string: L2=1.6 L_2 = 1.6 ft, thickness t2=332 t_2 = \frac{3}{32} inch
The frequency ratio is:
f2f1=L1L2×μ1μ2\frac{f_2}{f_1} = \frac{L_1}{L_2} \times \sqrt{\frac{\mu_1}{\mu_2}}
Since μ∝thickness2 \mu \propto \text{thickness}^2 :
μ1μ2=(t1t2)2=(1/323/32)2=(13)2=19\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:
f2f1=2.01.6×19=54×13=512\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, f2=512f f_2 = \frac{5}{12} f .

STEP 5

For one-third tension:
The frequency ratio considering tension T2=13T1 T_2 = \frac{1}{3} T_1 :
f2f1=L1L2×μ1μ2×T2T1=512×13\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}}
=512×13=5123= \frac{5}{12} \times \frac{1}{\sqrt{3}} = \frac{5}{12\sqrt{3}}
So, f2=5123f f_2 = \frac{5}{12\sqrt{3}} f .
The frequencies are: (a) 512f \frac{5}{12} f under equal tension. (b) 5123f \frac{5}{12\sqrt{3}} f under one-third tension.

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