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Chapter 10: Problem 1478

The tension in a wire is decreased by \(19 \%\), then the percentage decrease in frequency will be....... (A) \(19 \%\) (B) \(10 \%\) (C) \(0.19 \%\) (D) None of these

Short Answer

Expert verified
The percentage decrease in frequency when the tension in a wire is decreased by \(19\%\) is approximately \(10\%\).

Step by step solution

01

Recall the relationship between tension and frequency in a wire

According to the wave equation for a stretched wire, the frequency of a vibrating wire depends on the tension (F), the mass per unit length (µ), and the length of the wire (L). Mathematically, this can be represented as: \(f = \dfrac{1}{2L} \sqrt{\dfrac{F}{µ}}\) Where: - f is the frequency of the wire - F is the tension in the wire - µ is the mass per unit length - L is the length of the wire
02

Determine the new frequency

Let the initial frequency be \(f_1\) and the initial tension be \(F_1\). When the tension decreases by 19%, the new tension \(F_2\) can be found as: \(F_2 = F_1 - 0.19F_1 = 0.81F_1\) Now, let's find the new frequency \(f_2\) when the tension is F_2: \(f_2 = \dfrac{1}{2L} \sqrt{\dfrac{F_2}{µ}} = \dfrac{1}{2L} \sqrt{\dfrac{0.81F_1}{µ}}\)
03

Calculate the percentage decrease in frequency

To find the percentage decrease in frequency, we first need to find the ratio of the new frequency to the initial frequency: Percentage decrease = \(\dfrac{f_1 - f_2}{f_1} \times 100\%\) Substitute \(f_1\) and \(f_2\) from the earlier equations: Percentage decrease = \(\dfrac{\dfrac{1}{2L} \sqrt{\dfrac{F_1}{µ}} - \dfrac{1}{2L} \sqrt{\dfrac{0.81F_1}{µ}}}{\dfrac{1}{2L}\sqrt{\dfrac{F_1}{µ}}} \times 100\% \) Simplify the equation: Percentage decrease = \(\dfrac{\sqrt{\dfrac{F_1}{µ}} - \sqrt{\dfrac{0.81F_1}{µ}}}{\sqrt{\dfrac{F_1}{µ}}} \times 100\% \) Percentage decrease = \(1 - \sqrt{0.81} \times 100\% \) Percentage decrease ≈ 10% Therefore, the correct option is: (B) 10%

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Most popular questions from this chapter

A wire stretched between two rigid supports vibrates with a frequency of $45 \mathrm{~Hz}\(. If the mass of the wire is \)3.5 \times 10^{-2} \mathrm{~kg}$ and its linear mass density is \(4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\), what will be the tension in the wire? (A) \(212 \mathrm{~N}\) (B) \(236 \mathrm{~N}\) (C) \(248 \mathrm{~N}\) (D) \(254 \mathrm{~N}\)

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

An open organ pipe has fundamental frequency \(100 \mathrm{~Hz}\). What frequency will be produced if its one end is closed? (A) \(100,200,300, \ldots\) (B) \(50,150,250 \ldots .\) (C) \(50,100,200,300 \ldots \ldots\) (D) \(50,100,150,200 \ldots \ldots\)

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The frequency of the particle is \(\ldots \ldots \mathrm{s}^{-1}\). (A) \((1 / \pi)\) (B) \(\pi\) (C) \((1 / 2 \pi)\) (D) \((\pi / 2)\)

A wire of length \(10 \mathrm{~m}\) and mass \(3 \mathrm{~kg}\) is suspended from a rigid support. The wire has uniform cross sectional area. Now a block of mass \(1 \mathrm{~kg}\) is suspended at the free end of the wire and a wave having wavelength \(0.05 \mathrm{~m}\) is produced at the lower end of the wire. What will be the wavelength of this wave when it reached the upper end of the wire? (A) \(0.12 \mathrm{~m}\) (B) \(0.18 \mathrm{~m}\) (C) \(0.14 \mathrm{~m}\) (D) \(0.10 \mathrm{~m}\)
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