/*! 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 146 A string of mass \(2.5 \mathrm{~... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 146

A string of mass \(2.5 \mathrm{~kg}\) is under tension of \(200 \mathrm{~N}\). The length of the stretched string is \(20.0 \mathrm{~m}\). If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in (A) \(1 \mathrm{~s}\) (B) \(0.5 \mathrm{~s}\) (C) \(2 \mathrm{~s}\) (D) Data given is insufficient

Short Answer

Expert verified
The time taken for the disturbance to travel the length of the string is 0.5 seconds. To find this, first calculate the linear mass density (\(\mu\)) of the string as \(\frac{2.5\,\mathrm{kg}}{20.0\,\mathrm{m}} = 0.125\,\mathrm{kg/m}\). Then, determine the wave speed on the string using the formula \(v = \sqrt{\frac{T}{\mu}}\), which yields a wave speed of 40 m/s. Finally, calculate the time taken using the formula \(t = \frac{d}{v}\), resulting in 0.5 seconds (Option B).

Step by step solution

01

Calculate the linear mass density of the string

To find the linear mass density (\(\mu\)) of the string, divide the mass (\(m\)) of the string by its length (\(L\)): \(\mu = \frac{m}{L}\) Plugging in the values for mass (2.5 kg) and length (20.0 m): \(\mu = \frac{2.5\,\mathrm{kg}}{20.0\,\mathrm{m}} = 0.125\,\mathrm{kg/m}\).
02

Calculate the speed of the wave on the string

Using the formula for wave speed: \(v = \sqrt{\frac{T}{\mu}}\) where \(T\) is the tension (200 N) and \(\mu\) is the linear mass density (0.125 kg/m), we can find the wave speed: \(v = \sqrt{\frac{200\,\mathrm{N}}{0.125\,\mathrm{kg/m}}} = 40\,\mathrm{m/s}\).
03

Calculate the time taken for the disturbance to travel the length of the string

Finally, find the time taken (\(t\)) for the disturbance to travel the length of the string using the formula: \(t = \frac{d}{v}\) where \(d\) is the distance (20.0 m) and \(v\) is the wave speed (40 m/s): \(t = \frac{20.0\,\mathrm{m}}{40\,\mathrm{m/s}} = 0.5\,\mathrm{s}\). Thus, the disturbance will reach the other end in 0.5 seconds, which corresponds to Option (B).

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 91影视!

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

Two monochromatic coherent point sources \(S_{I}\) and \(S_{2}\) are separated by a distance \(L .\) Each source emits light of wavelength \(\lambda\), where \(L \gg \lambda\). The line \(S_{1} S_{2}\) when extended meets a screen perpendicular to it at a point \(A\). (A) The interference fringes on the screen are circular in shape. (B) The interference fringes on the screen are straight lines perpendicular to the line \(S_{1} S_{2} A\). (C) The point \(A\) is an intensity maxima if \(L=n \lambda\). (D) The point \(A\) is always an intensity maxima for any separation \(L\).

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) \(\frac{2}{b}\)

Two trains \((A\) and \(B\) ) are moving towards each other on two parallel tracks at the same speed with respect to the ground. The whistle of train \(A\) blows. In which of the following cases, the frequency of the sound heard by a passenger on the other train \(B\) will be greatest? (A) If the air is still. (B) If a wind blows in the same direction and at the same speed as the other train \(B\). (C) If a wind blows in the opposite direction and at the same speed as the other train \(B\). (D) Frequency will be same in the above three cases.

The area of region covered by the TV broadcast by a TV tower of \(100 \mathrm{~m}\) height will be (radius of the earth \(=6.4 \times 10^{6} \mathrm{~m}\) ) (A) \(12.8 \pi \times 10^{8} \mathrm{~km}^{2}\) (B) \(1.28 \pi \times 10^{3} \mathrm{~km}^{2}\) (C) \(0.64 \pi \times 10^{3} \mathrm{~km}^{2}\) (D) \(1.28 \times 10^{3} \mathrm{~km}^{2}\)

A mass \(M\) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period \(T\). If the mass is increased by \(m\), the time period becomes \(\frac{5 T}{3}\). Then the radio of \(\frac{m}{M}\) is (A) \(\frac{3}{5}\) (B) \(\frac{25}{9}\) (C) \(\frac{16}{9}\) (D) \(\frac{5}{3}\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Classical Mechanics

Read Explanation

Turning Points in Physics

Read Explanation

Kinematics Physics

Read Explanation

Waves Physics

Read Explanation

Quantum Physics

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.