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Chapter 10: Problem 1464

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) \(254 \mathrm{~N}\)

Short Answer

Expert verified
The tension in the wire is approximately \(248 N\). The correct answer is (C) \(248 \mathrm{~N}\).

Step by step solution

01

Write the formula for the fundamental frequency of a vibrating wire.

The formula for the fundamental frequency of a vibrating wire is given by: \[f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\] where: - \(f\) is the frequency of the wire (\(45 Hz\)), - \(L\) is the length of the wire, which we need to find, - \(T\) is the tension in the wire, which is what we want to determine, - and \(\mu\) is the linear mass density of the wire (\(4.0 \times 10^{-2} kg/m\)).
02

Find the length of the wire using the mass and linear mass density.

We can find the length of the wire using the given mass and linear mass density. The linear mass density is defined as the mass per unit length of the wire: \[\mu = \frac{m}{L}\] where: - \(\mu\) is the linear mass density (\(4.0 \times 10^{-2} kg/m\)), - \(m\) is the mass of the wire (\(3.5 \times 10^{-2} kg\)), - and \(L\) is the length of the wire. We can rearrange this equation to solve for the length: \[L = \frac{m}{\mu} = \frac{3.5 \times 10^{-2} kg}{4.0 \times 10^{-2} kg/m} = 0.875 m\]
03

Rearrange the formula of the fundamental frequency to solve for the tension.

We can rearrange the formula of the fundamental frequency to solve for the tension: \[T = \mu (2Lf)^2\]
04

Plug in the values and calculate the tension in the wire.

Now, we can plug in the values for the linear mass density, length, and frequency into the equation and calculate the tension in the wire: \[T = (4.0 \times 10^{-2} kg/m)(2 \times 0.875 m \times 45 Hz)^2\] \[T = (4.0 \times 10^{-2} kg/m)(157.5 Hz)^2\] \[T \approx 248 N\] The tension in the wire is approximately \(248 N\). The correct answer is (C) \(248 \mathrm{~N}\).

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Most popular questions from this chapter

If the kinetic energy of a particle executing S.H.M. is given by \(\mathrm{K}=\mathrm{K}_{0} \cos ^{2} \omega \mathrm{t}\), then the displacement of the particle is given by \(\ldots \ldots\) (A) $\left\\{\mathrm{K}_{0} / \mathrm{m} \omega^{2}\right\\} \sin \omega \mathrm{t}$ (B) $\left\\{\left(2 \mathrm{~K}_{0}\right) /\left(\mathrm{m} \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t$ (C) \(\left\\{2 \omega^{2} / \mathrm{mK}_{0}\right\\} \sin \omega t\) (D) $\left\\{2 \mathrm{~K}_{0} / \mathrm{m} \omega\right\\}^{1 / 2} \sin \omega t$

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