Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 54

A string of length \(1.5 \mathrm{~m}\) with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is \(4 \mathrm{~mm}\). Minimum distance between the two points having amplitude \(2 \mathrm{~mm}\) is : (1) \(1 \mathrm{~m}\) (2) \(75 \mathrm{~cm}\) (3) \(60 \mathrm{~cm}\) (4) \(50 \mathrm{~cm}\)

Short Answer

Expert verified
1 m

Step by step solution

01

Understand the Fundamental Mode

In the fundamental mode of vibration for a string with both ends clamped, there is one antinode at the center and nodes at the two ends. The length of the string \( L \) is equal to half the wavelength \( \frac{\lambda}{2} \). Thus, \( \lambda = 2L = 2 \times 1.5 \text{ m} = 3 \text{ m} \).
02

Identify Amplitudes and Distances

Amplitude in a standing wave depends on the position along the string. The maximum amplitude at the antinode (center of the string) is 4 mm.
03

Use Sine Function for Amplitude

The amplitude \( A(x) \) at a distance \( x \) from a node is given by the function: \[ A(x) = A_{\text{max}} \sin \left( \frac{\pi x}{L} \right) \] where \( A_{\text{max}} \) is the maximum amplitude (4 mm).
04

Set Up the Equation for Given Amplitude

For the points where the amplitude is 2 mm, use \[ A(x) = 2 = 4 \sin \left( \frac{\pi x}{1.5} \right) \] Simplifying, \[ \sin \left( \frac{\pi x}{1.5} \right) = \frac{1}{2} \]
05

Solve for \( x \)

\( \sin \left( \frac{\pi x}{1.5} \right) = \frac{1}{2} \) has solutions \( \frac{\pi x}{1.5} = \frac{\pi}{6} \) and \( \frac{\pi x}{1.5} = \frac{5\pi}{6} \). Thus, \( x = \frac{1.5}{6} = 0.25 \text{ m} \) and \( x = 1.25 \text{ m} \).
06

Calculate the Minimum Distance

The distance between these two points is \( x_2 - x_1 = 1.25 \text{ m} - 0.25 \text{ m} = 1.00 \text{ m} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standing Waves
Standing waves are a fascinating phenomenon seen in many physical systems, such as musical instruments and vibrating strings. They occur when two waves of the same frequency and amplitude travel in opposite directions and interfere with one another. This interference results in a wave pattern that appears to be standing still. In the case of a string clamped at both ends, standing waves are produced as the wave reflects back and forth along the string, forming fixed points called nodes and antinodes. Nodes are points where the wave has zero amplitude and remain stationary, while antinodes are points where the wave reaches maximum amplitude and undergoes maximum oscillation.
Amplitude
Amplitude is a key property of waves that describes the maximum displacement of points on the wave from their rest position. It determines the energy and intensity of the wave. In a standing wave on a vibrating string, the amplitude varies along the length of the string. The maximum amplitude, known as the antinode, is located at the center when the string vibrates in its fundamental mode. In the given exercise, the center of the string has an amplitude of 4 mm, while points along the string will have lower amplitudes depending on their position relative to the nodes and antinodes.
Nodal Points
Nodal points, or simply nodes, are specific points on a standing wave where there is no movement. This means the amplitude is zero due to the destructive interference of the two waves traveling in opposite directions. On a string clamped at both ends, these nodes appear at the endpoints and at various intervals along the string. The distance between two consecutive nodes is half the wavelength of the wave. When dealing with the fundamental mode of a vibrating string, the nodes are found at the two ends of the string and any additional nodes that form depend on the harmonic mode of the vibration.
Wavelength
Wavelength is the distance between successive crests, troughs, or identical points of the wave. It plays a critical role in determining the frequency and speed of wave propagation. For a string clamped at both ends vibrating in its fundamental mode, the wavelength is twice the length of the string. In the exercise, the string's length is 1.5 meters, so the wavelength is 3 meters. By understanding the wavelength, we can predict where nodes and antinodes will form along the string. Moreover, wavelength helps us solve for points with specific amplitudes by using mathematical relationships, such as the sine function, to describe wave behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The number of real points at which line \(3 y-2 x=14\) cuts hyperbola \(\frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{4}=5\) is (1) 0 (2) 1 (3) 2 (4) 4

In how many of the following compounds of sulphur, there is S-S bond (only single bond not double bond between two sulphur atoms). \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{4}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{5}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\), \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}, \mathrm{~S}_{2} \mathrm{Cl}_{2}\), cyclic \(\left(\mathrm{SO}_{3}\right)_{3}\) (1) 7 (2) 6 (3) 5 (4) 4

Moment of inertia of a uniform quarter disc of radius \(\mathrm{R}\) and mass \(\mathrm{M}\) about an axis through its centre of mass and perpendicular to its plane is : (1) \(\frac{\mathrm{MR}^{2}}{2}-\mathrm{M}\left(\frac{4 \mathrm{R}}{3 \pi}\right)^{2}\) (2) \(\frac{\mathrm{MR}^{2}}{2}-\mathrm{M}\left(\sqrt{2} \frac{4 \mathrm{R}}{3 \pi}\right)^{2}\) (3) \(\frac{\mathrm{MR}^{2}}{2}+\mathrm{M}\left(\frac{4 \mathrm{R}}{3 \pi}\right)^{2}\) (4) \(\frac{\mathrm{MR}^{2}}{2}+\mathrm{M}\left(\sqrt{2} \frac{4 \mathrm{R}}{3 \pi}\right)^{2}\)

The angle of elevation of the top of a tower standing on a horizontal plane from a point \(A\) is \(\alpha\). After walking a distance \(d\) towards the foot of the tower the angle of elevation is found to be \(\beta\). The height of the tower is- (1) \(\frac{d}{\cot \alpha+\cot \beta}\) (2) \(\frac{d}{\cot \alpha-\cot \beta}\) (3) \(\frac{d}{\tan \beta-\tan \alpha}\) (4) \(\frac{d}{\tan \beta+\tan \alpha}\)

If the ratio of the intensity of two coherent sources is 4 then the visibility \(\left[\left(I_{\max }-I_{\min }\right) /\left(I_{\max }+I_{\min }\right)\right]\) of the fringes is - (1) 4 (2) \(4 / 5\) (3) \(3 / 5\) (4) 9
See all solutions

Recommended explanations on English Textbooks

English Language Study

Read Explanation

Free Response Essay

Read Explanation

International English

Read Explanation

English Grammar

Read Explanation

Single Paragraph Essay

Read Explanation

Essay Plan

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free