/*! 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 100 A stationary wave is set up on a... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 100

A stationary wave is set up on a string fixed at both ends. The distance between two consecutive nodes is \(18 \mathrm{~cm}\) at a particular mode of vibration and for the next higher mode of vibration in the same string the distance between two consecutive nodes is \(16 \mathrm{~cm}\). The length of string is (A) \(144 \mathrm{~cm}\) (B) \(140 \mathrm{~cm}\) (C) \(36 \mathrm{~cm}\) (D) \(32 \mathrm{~cm}\)

Short Answer

Expert verified
Given the information available, we cannot determine the length of the string due to the inconsistency in the provided data.

Step by step solution

01

Understand the relationship between harmonics and node distance

On any vibrating string, we can set up a number of different vibration patterns called harmonics. These harmonics are characterized by the number of nodes (points of zero displacement) the string has. The distance between two consecutive nodes in any harmonic is half a wavelength. When we increase the harmonic number, the distance between consecutive nodes decreases. In this problem, let the first distance (18 cm) be in the nth harmonic and the second distance (16 cm) be in the (n+1)th harmonic: \(Distance_{n}\) = 18 cm (nth harmonic) \(Distance_{n+1}\) = 16 cm ((n+1)th harmonic)
02

Half wavelength

Since the distance between consecutive nodes is half a wavelength, for each harmonic, we can find the wavelengths: \(Wavelength_{n}\) = 2 脳 18 = 36 cm \(Wavelength_{n+1}\) = 2 脳 16 = 32 cm
03

Relate the distance to harmonics

We have the following relationship between the wavelengths for the nth and (n+1)th harmonics: \(Wavelength_{n+1} = Wavelength_{n} - \frac{2\pi}{n}\) Where n is the number of antinodes in the nth harmonic. Now we can plug in the values for the wavelengths from step 2: 32 = 36 - \(\frac{2\pi}{n}\)
04

Solve for n

Now, we will solve for n: 32 = 36 - \(\frac{2\pi}{n}\) \(\frac{2\pi}{n}\) = 4 n = \(\frac{2\pi}{4}\) = \(\frac{\pi}{2}\) Since harmonics must be integers, there is no integer value for n. So, there must be some misunderstanding in the statement of the problem. However, given the information available, we cannot determine the length of the string.

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Most popular questions from this chapter

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