/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A sonometer wire of density \(d\... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 22

A sonometer wire of density \(d\) and radius \(r\) is held between two bridges at a distance \(L\) apart. The wire has a tension \(T\). The fundamental frequency of the wire will be (A) \(f=\frac{1}{2 L r} \sqrt{\frac{T}{\pi d}}\) (B) \(f=\frac{r}{2 L} \sqrt{\frac{\pi d}{T}}\) (C) \(f=\frac{1}{2 L r} \sqrt{\frac{d}{\pi T}}\) (D) \(f=\frac{1}{2 L} \sqrt{\frac{d}{T}}\)

Short Answer

Expert verified
\(f = \frac{1}{2L} \sqrt{\frac{T}{\pi d r^2}}\)

Step by step solution

01

Find the expression for linear mass density (\(\mu\))

The linear mass density (\(\mu\)) can be defined as the mass per unit length. In this case, we are given the density (\(d\)) and the radius (\(r\)) of the wire. The volume of the wire can be given as: \(V = \pi r^2 L\), where \(\pi r^2\) is the cross-sectional area of the wire. The mass of the wire can be calculated using this volume and the density (\(d\)): \(m = d \cdot V = d \cdot (\pi r^2 L)\). Since linear mass density is the mass per unit length, we can write \(\mu\) as: \(\mu = \frac{m}{L} = d \cdot (\pi r^2)\). Now that we have an expression for the linear mass density in terms of the given variables, we can substitute it into the formula for the fundamental frequency.
02

Calculate the fundamental frequency using the formula

We know the formula for the fundamental frequency is: \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) Now, substitute the expression for \(\mu\) from Step 1: \(f = \frac{1}{2L}\sqrt{\frac{T}{d \cdot (\pi r^2)}}\) Simplifying the expression, we get: \(f = \frac{1}{2L} \sqrt{\frac{T}{\pi d r^2}}\) Comparing this expression to the given options, we see that the correct option is: (A) \(f = \frac{1}{2L} \sqrt{\frac{T}{\pi d r^2}}\)

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Most popular questions from this chapter

A stationary source of sound is emitting waves of frequency \(30 \mathrm{~Hz}\) towards a stationary wall. There is an observer standing between the source and the wall. If the wind blows from the source to the wall with a speed \(30 \mathrm{~m} / \mathrm{s}\) then the number of beats heard by the observer is (velocity of sound with respect to wind is \(330 \mathrm{~m} / \mathrm{s})\) (A) 10 (B) 3 (C) 6 (D) Zero

The amplitude of a damped oscillator decreases to \(0.9\) times its original magnitude in \(5 \mathrm{~s}\). In another \(10 \mathrm{~s}\) it will decrease to \(\alpha\) times its original magnitude, where \(\alpha\) equals (A) \(0.81\) (B) \(0.729\) (C) \(0.6\) (D) \(0.7\)

If a simple pendulum has significant amplitude (up to a factor \(1 / e\) of original) only in the period between \(t=0 s\) to \(t=\tau s\), then \(\tau\) may be called the average lift of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with \(b\) as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds: (A) \(\frac{0.693}{b}\) (B) \(b\) (C) \(\frac{1}{b}\) (D) \(\frac{2}{b}\)

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An open glass tube is immersed in mercury in such a way that a length of \(8 \mathrm{~cm}\) extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional \(46 \mathrm{~cm}\). What will be length of the air column above mercury in the tube now? (Atmospheric pressure \(=76 \mathrm{~cm}\) of \(\mathrm{Hg}\) ) (A) \(16 \mathrm{~cm}\) (B) \(22 \mathrm{~cm}\) (C) \(38 \mathrm{~cm}\) (D) \(6 \mathrm{~cm}\)
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