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Chapter 15: Problem 38

A string is \(35.0 \mathrm{~cm}\) long and has a mass per unit length of \(5.51 \cdot 10^{-4} \mathrm{~kg} / \mathrm{m}\). What tension must be applied to the string so that it vibrates at the fundamental frequency of \(660 \mathrm{Hz?}\)

Short Answer

Expert verified
Answer: The tension that must be applied to the string is approximately 56.38 N.

Step by step solution

01

Ensure all values are in the correct units

Since the length of the string, \(L\) is given in cm, we need to convert it into meters. The mass per unit length is given in kg/m, which is already in the correct units. \(L = 35.0 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.35 \mathrm{~m}\) Now we have all the values we need: Length of the string, L = 0.35 m Mass per unit length, μ = \(5.51 \times 10^{-4} \mathrm{~kg/m}\) Fundamental frequency, \(f_1 = 660 \mathrm{~Hz}\)
02

Solve for tension T using the formula

We're solving for \(T\) using the formula: $$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$ First, we'll isolate T by squaring both sides of the equation and then multiply by the denominator: $$T = \mu(2Lf_1)^2$$ Next, we'll plug in the values we have for L, μ, and \(f_1\): $$T = (5.51 \times 10^{-4} \mathrm{~kg/m})\left(2\times(0.35 \mathrm{~m})\times(660 \mathrm{~Hz})\right)^2$$
03

Calculate the tension T

Now, we can perform the calculations: $$T = (5.51 \times 10^{-4} \mathrm{~kg/m}) \times (4\times0.35 \mathrm{~m}\times660 \mathrm{~Hz})^2$$ $$T ≈ 56.38 \mathrm{~N}$$ This gives the tension in the string that must be applied for it to vibrate at the fundamental frequency of 660 Hz.

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Most popular questions from this chapter

Consider a monochromatic wave on a string, with amplitude \(A\) and wavelength \(\lambda\), traveling in one direction. Find the relationship between the maximum speed of any portion of string, \(v_{\max },\) and the wave speed, \(v\)

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(T=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different frequencies \((\omega=3000\). rad/s and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s}\) ) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

a) Starting from the general wave equation (equation 15.9 ), prove through direct derivation that the Gaussian wave packet described by the equation \(y(x, t)=(5.00 m) e^{-0.1(x-5 t)^{2}}\) is indeed a traveling wave (that it satisfies the differential wave equation). b) If \(x\) is specified in meters and \(t\) in seconds, determine the speed of this wave. On a single graph, plot this wave as a function of \(x\) at \(t=0, t=1.00 \mathrm{~s}, t=2.00 \mathrm{~s},\) and \(t=3.00 \mathrm{~s}\) c) More generally, prove that any function \(f(x, t)\) that depends on \(x\) and \(t\) through a combined variable \(x \pm v t\) is a solution of the wave equation, irrespective of the specific form of the function \(f\)

A traveling wave propagating on a string is described by the following equation: $$ y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$ a) Determine the minimum separation, \(\Delta x_{\min }\), between two points on the string that oscillate in perfect opposition of phases (move in opposite directions at all times). b) Determine the separation, \(\Delta x_{A B}\), between two points \(A\) and \(B\) on the string, if point \(B\) oscillates with a phase difference of 0.7854 rad compared to point \(A\). c) Find the number of crests of the wave that pass through point \(A\) in a time interval \(\Delta t=10.0 \mathrm{~s}\) and the number of troughs that pass through point \(B\) in the same interval. d) At what point along its trajectory should a linear driver connected to one end of the string at \(x=0\) start its oscillation to generate this sinusoidal traveling wave on the string?

The speed of light waves in air is greater than the speed of sound in air by about a factor of a million. Given a sound wave and a light wave of the same wavelength, both traveling through air, which statement about their frequencies is true? a) The frequency of the sound wave will be about a million times greater than that of the light wave. b) The frequency of the sound wave will be about a thousand times greater than that of the light wave. c) The frequency of the light wave will be about a thousand times greater than that of the sound wave. d) The frequency of the light wave will be about a million times greater than that of the sound wave. e) There is insufficient information to determine the relationship between the two frequencies.
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