Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 1476

A string of linear density \(0.2 \mathrm{~kg} / \mathrm{m}\) is stretched with a force of \(500 \mathrm{~N}\). A transverse wave of length \(4.0 \mathrm{~m}\) and amplitude \(1 / 1\) meter is travelling along the string. The speed of the wave is \(\ldots \ldots \ldots \ldots \mathrm{m} / \mathrm{s}\) (A) 50 (B) \(62.5\) (C) 2500 (D) \(12.5\)

Short Answer

Expert verified
The speed of the wave is \(50 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Recall the formula for wave speed

We use the formula for wave speed on a string, given by: \(v = \sqrt{\frac{T}{\mu}}\) where \(v\) is the wave speed, \(T\) is the tension force, and \(\mu\) is the linear density of the string.
02

Plug in the given values

We are given the linear density \(\mu = 0.2 \mathrm{~kg} / \mathrm{m}\) and the tension force \(T = 500 \mathrm{~N}\). We can plug these values into the formula to find the wave speed: \(v = \sqrt{\frac{500 \mathrm{~N}}{0.2 \mathrm{~kg} / \mathrm{m}}}\)
03

Solve for the wave speed

Now, we can perform the calculation to find the wave speed: \(v = \sqrt{\frac{500 \mathrm{~N}}{0.2 \mathrm{~kg} / \mathrm{m}}} = \sqrt{2500 \mathrm{~m} / \mathrm{s}} = 50 \mathrm{~m} / \mathrm{s}\)
04

Compare the result with the given choices

The wave speed we calculated is \(50 \mathrm{~m} / \mathrm{s}\), which corresponds to choice (A). So the correct answer is: (A) 50

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!

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

A system is executing S.H.M. with a periodic time of \(4 / 5 \mathrm{~s}\) under the influence of force \(\mathrm{F}_{1}\) When a force \(\mathrm{F}_{2}\) is applied, the periodic time is \((2 / 5) \mathrm{s}\). Now if \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) are applied simultaneously along the same direction, the periodic time will be......... (A) \(\\{4 /(5 \sqrt{5})\\}\) (B) \(\\{4 /(4 \sqrt{5})\\}\) (C) \(\\{8 /(4 \sqrt{5})\\}\) (D) \(\\{8 /(5 \sqrt{5})\\}\)

One end of a mass less spring having force constant \(\mathrm{k}\) and length \(50 \mathrm{~cm}\) is attached at the upper end of a plane inclined at an angle \(e=30^{\circ} .\) When a body of mass \(m=1.5 \mathrm{~kg}\) is attached at the lower end of the spring, the length of the spring increases by $2.5 \mathrm{~cm}$. Now, if the mass is displaced by a small amount and released, the amplitude of the resultant oscillation is ......... (A) \((\pi / 7)\) (B) \((2 \pi / 7)\) (C) \((\pi / 5)\) (D) \((2 \pi / 5)\)

When a mass \(\mathrm{m}\) is suspended from the free end of a massless spring having force constant \(\mathrm{k}\), its oscillates with frequency \(\mathrm{f}\). Now if the spring is divided into two equal parts and a mass $2 \mathrm{~m}$ is suspended from the end of anyone of them, it will oscillate with a frequency equal to......... (A) \(\mathrm{f}\) (B) \(2 \mathrm{f}\) (C) \((\mathrm{f} / \sqrt{2})\) (D) \(\sqrt{2 f}\)

For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a) Statement \(-1\) is true, statement \(-2\) is true; statement \(-2\) is the correct explanation of statement \(-1\) (b) Statement \(-1\) is true, statement \(-2\) is true but statement \(-2\) is not the correct explanation of statement \(-1\). (c) Statement \(-1\) is true, statement \(-2\) is false (d) Statement \(-1\) is false, statement \(-2\) is true (A) a (B) b (C) \(\mathrm{c}\) (D) \(\mathrm{d}\) Statement \(-1:\) If the length of a simple pendulum is increased by \(3 \%\), then the periodic time changes by \(1.5 \%\). Statement \(-2:\) Periodic time of a simple pendulum is proportional to its length. (A) a (B) \(b\) (C) \(c\) (D) d

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of frequencies \((\mathrm{f} / \mathrm{f})=\ldots \ldots \ldots\) (A) \(\sqrt{\\{} \mathrm{M} /(\mathrm{m}+\mathrm{M})\\}\) (B) \(\sqrt{\\{\mathrm{m} /(\mathrm{m}+\mathrm{M})\\}}\) (C) \(\sqrt{\\{\mathrm{MA} / \mathrm{mA}}\\}\) (D) \(\sqrt{[}\\{(\mathrm{M}+\mathrm{m}) \mathrm{A}\\} / \mathrm{mA}]\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

Pragmatics

Read Explanation

Syntax

Read Explanation

Phonetics

Read Explanation

Email

Read Explanation

Argumentative Essay

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