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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

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Key Concepts

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

Linear Density
Linear density is a measure of mass per unit length. In the context of waves on a string, it tells us how much mass is distributed along each meter of the string. It is often represented by the Greek letter \( \mu \). For example, if a string has a linear density of \( 0.2 \mathrm{~kg} / \mathrm{m} \), it means that every meter of the string weighs 0.2 kg.
Understanding linear density is crucial when calculating the wave speed, as it directly affects how the wave travels through the string. A higher linear density means that the string has more mass, which can slow down the wave speed. Conversely, a lower linear density suggests less mass and potentially a faster wave. Knowing this helps choose the right parameters for applications that involve waves on strings, such as musical instruments.
To calculate linear density, you can divide the total mass of the string by its total length, i.e., \( \mu = \frac{m}{L} \), where \( m \) is the total mass and \( L \) is the length of the string.
Tension in String
Tension in a string refers to the force that is applied along the length of the string, keeping it taut. In many mechanical systems, tension is crucial for maintaining the stability and functionality of components involving strings or cables. The tension force in our example is given as \(500 \mathrm{~N}\), which is necessary for calculating the wave speed.
When a string has high tension, it becomes tighter and waves can travel faster through it. Conversely, if the tension is low, the string is looser and waves may travel slower. Think about the strings of a guitar or violin; tuning these strings changes their tension and thereby affects the pitch of the sound we hear.
Tension is often adjusted by increasing or decreasing the force applied to the string's ends, such as turning the pegs on a guitar.
Transverse Wave
A transverse wave is a type of wave where the motion of the medium's particles is perpendicular to the direction of the wave's travel. Waves on a string, like the ones in musical instruments, are usually transverse waves. This means as the wave travels horizontally along a string, the particles of the string move up and down.
In the given exercise, the wave has an amplitude—a measure of its height from the center line to a peak—of 1 meter. The amplitude of a wave affects its energy; higher amplitude means higher energy.
In conciseness, think of a transverse wave as ripples on a pond when you throw a stone in the water. Energy travels outward in perpendicular oscillations compared to the direction of wave travel. These kinds of waves show how energy can transfer through a medium without the whole medium traveling with the wave.
Wave Speed Formula
The wave speed formula is essential for determining how quickly a wave travels through a medium. When dealing with waves on a string, the wave speed \(v\) can be calculated using the formula: \[v = \sqrt{\frac{T}{\mu}}\] where \(T\) represents the tension in the string and \(\mu\) is its linear density.
This formula illustrates that the wave speed is directly affected by the tension in the string and inversely affected by its linear density. More tension allows the wave to move faster, while higher linear density makes it slower. So, to increase wave speed in a string, you can either increase the tension, decrease the linear density, or both.
The formula plays a prime role in various applications, such as designing strings for musical instruments, where specific speeds are crucial for producing the desired pitch or tone.

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Most popular questions from this chapter

A rocket is moving at a speed of \(130 \mathrm{~m} / \mathrm{s}\) towards a stationary target. While moving, it emits a wave of frequency \(800 \mathrm{~Hz}\). Calculate the frequency of the sound as detected by the target. (Speed of wave \(=330 \mathrm{~m} / \mathrm{s}\) ) (A) \(1320 \mathrm{~Hz}\) (B) \(2540 \mathrm{~Hz}\) (C) \(1270 \mathrm{~Hz}\) (D) \(660 \mathrm{~Hz}\)

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) \(\\{8 /(5 \sqrt{5})\\}\)

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of \((\mathrm{g} / 3)\), then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is \((\pi / 10) \mathrm{s}\), then the maximum force acting on body is \(\ldots \ldots \ldots \ldots\) (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

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