/*! 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 123 For a particular mode of vibrati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For a particular mode of vibration of string, the distance between two consecutive nodes is \(18 \mathrm{~cm}\). For the next higher mode, the distance becomes \(16 \mathrm{~cm}\). The length of the string is (A) \(18 \mathrm{~cm}\) (B) \(16 \mathrm{~cm}\) (C) \(144 \mathrm{~cm}\) (D) \(72 \mathrm{~cm}\)

Short Answer

Expert verified
The length of the string is 144 cm, so the correct answer is (C) 144 cm.

Step by step solution

01

Derive the relationship between consecutive distances and length of the string

If we consider the nth mode, the length of the string L is given by: L = (n / 2) λ_n , where λ_n is the wavelength of the nth mode of vibration. Similarly, for the (n+1)th mode: L = ((n + 1) / 2) λ_(n+1). Since we are given the distances between consecutive nodes for two different modes, we can set up an equation relating the string length (L) and the distance between the nodes for each mode.
02

Set up the equation based on given distances

The distance between two consecutive nodes for the nth mode is given as 18 cm, which is equal to half the wavelength (λ_n): λ_n = 36 cm The distance between two consecutive nodes for the (n+1)th mode is given as 16 cm, which is equal to half the wavelength (λ_(n+1)): λ_(n+1) = 32 cm Now, plug in the values of λ_n and λ_(n+1) into the equations for the string length L: L = (n / 2) (36 cm) L = ((n + 1) / 2) (32 cm)
03

Solve for n

Now we have two equations with the same unknown, L. We can set them equal to each other and solve for n: ( n / 2) (36 cm) = ((n + 1) / 2) (32 cm) To eliminate the fractions, multiply both sides by 2: n (36 cm) = (n + 1) (32 cm) Distribute the expressions: 36n = 32n + 32 cm Subtract 32n from both sides: 4n = 32 cm Divide by 4: n = 8
04

Find the length of the string (L)

Now that we have the value of n, we can plug it back into one of the equations to find the length of the string, L. We can use the first equation: L = (n / 2) (36 cm) Plug in the value of n: L = (8 / 2) (36 cm) L = 4 (36 cm) Multiply to get the final answer: L = 144 cm The length of the string is 144 cm, so the correct answer is (C) 144 cm.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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

Which of the following statements are true for wave motion? (A) Mechanical transverse waves can propagate through all mediums. (B) Longitudinal waves can propagate through solids only. (C) Mechanical transverse waves can propagate through solids only. (D) Longitudinal waves can propagate through vacuum.

Assertion: There are two sound waves propagating in same medium having amplitudes and frequencies \(2 A\), \(f\) and \(A, 2 f\), respectively. The intensity of first wave is four times that of the other. Reason: Intensity of a wave \(I=\frac{1}{2} \rho v \omega^{2} A^{2}\) (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) D

A string of length \(l\) is fixed at both ends. It is vibrating in its third overtone. Maximum amplitude of the particles on the string is \(A\). The amplitude of the particle at a distance \(l / 3\) from one end is (A) \(A\) (B) 0 (C) \(\frac{\sqrt{3 A}}{2}\) (D) \(\frac{A}{2}\)

A long wire \(A B C\) is made by joining two wires \(A B\) and \(B C\) of equal area of cross-section. \(A B\) has length \(4.8 \mathrm{~m}\) and mass \(0.12 \mathrm{~kg}\) while \(B C\) has length \(2.56\) \(\mathrm{m}\) and mass \(0.4 \mathrm{~kg}\). The wire is under a tension of \(160 \mathrm{~N}\). A wave \(Y\) (in \(\mathrm{cm})=3.5 \sin (k x-w t)\) is sent along \(A B C\) from end \(A\). No power is dissipated during propagation of wave. Column-I (A) Amplitude of reflected wave (B) Amplitude of transmitted wave (C) Maximum displacement of antinodes in the wire \(A B\) (D) Percentage fraction of power transmitted in the wire \(B C\) \begin{tabular}{l} Column-II \\ \hline 1. \(2.0\) \end{tabular} \(1.5\) 2 3 . 5 \(-1\) 82 4 \(=\) 5 . 92

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) \(0.7\)

See all solutions

Recommended explanations on Physics Textbooks

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.