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Chapter 9: Problem 86

A stretched wire of some length under a tension is vibrating with its fundamental frequency. Its length is decreased by \(45 \%\) and tension is increased by \(21 \%\). Now its fundamental frequency (assuming linear mass density remains the same) (A) increases by \(50 \%\) (B) increases by \(100 \%\) (C) decreases by \(50 \%\) (D) decreases by \(25 \%\)

Short Answer

Expert verified
The new fundamental frequency increases by 100% when the length of the wire is decreased by 45% and the tension is increased by 21%. Thus, the correct answer is (B).

Step by step solution

01

Write down the formula for fundamental frequency.

The formula for the fundamental frequency of a stretched wire is given by: \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\) where: - \(f\) is the fundamental frequency, - \(L\) is the length of the wire, - \(T\) is the tension in the wire, and - \(\mu\) is the linear mass density.
02

Calculate the new length and tension.

Decrease the length by 45%: \(L_1 = L - 0.45L = 0.55L\) Increase the tension by 21%: \(T_1 = T + 0.21T = 1.21T\)
03

Calculate the new fundamental frequency.

Now substitute the new length and tension into the formula for the fundamental frequency to find the new fundamental frequency (\(f_1\)): \(f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}} = \frac{1}{2(0.55L)} \sqrt{\frac{1.21T}{\mu}}\)
04

Find the ratio of the new and initial fundamental frequencies.

Divide the new fundamental frequency by the initial one: \(\frac{f_1}{f} = \frac{\frac{1}{2(0.55L)} \sqrt{\frac{1.21T}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T}{\mu}}} = \frac{1.21}{0.55} = 2.2\)
05

Calculate the percentage increase in fundamental frequency.

Now, we can find the percentage increase in the fundamental frequency: Percentage increase = \(\frac{f_1 - f}{f} \times 100\% = \frac{2.2f - f}{f} \times 100\% = 1.2 \times 100\% = 120\%\) Now compare this result with the given options: (A) increases by 50% (B) increases by 100% (C) decreases by 50% (D) decreases by 25% Answer: (B) The new fundamental frequency increases by 100%.

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