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Statement-1: The speed of sound in solid is maximum though their density is large. Statement-2 : 'lhe coefficient of elasticity of solid is large.

Short Answer

Expert verified
Both statements are true, and Statement-2 explains Statement-1.

Step by step solution

01

Understand the Properties of Sound in Solids

Sound travels as a mechanical wave, which means it's dependent on the medium's properties such as elasticity and density. In solids, molecules are closely packed, making them dense systems.
02

Investigate the Impact of Elasticity

The speed of sound in a medium is influenced by its coefficient of elasticity (Y) and density (蟻). For solids, elasticity is typically very high, contributing significantly to high sound speeds.
03

Apply the Formula for Speed of Sound in Solids

Use the formula for the speed of sound in a medium: \( v = \sqrt{\frac{Y}{\rho}} \), where \( v \) is the speed of sound, \( Y \) is the Young's modulus (elasticity), and \( \rho \) is the density.
04

Analyze Statement-1

Statement-1 suggests sound speed is maximum in solids despite their density. Given the formula, high elasticity means a high Y/蟻 ratio, supporting the statement that sound speed can be high despite the large density due to even larger elasticity.
05

Analyze Statement-2

Statement-2, which claims that the coefficient of elasticity is large, aligns with the known properties of solids, where high elasticity corresponds to higher sound speeds. Thus, supporting Statement-1.
06

Correlate the Statements

By reviewing both statements, Statement-2 correctly explains why sound speed is highest in solids as addressed in Statement-1. High elasticity ensures a greater speed factor outweighing high density.

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Key Concepts

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

Mechanical Waves
Sound travels as a mechanical wave, which means it requires a medium to propagate. Mechanical waves differ from electromagnetic waves, like light, because they need particles to move through. In solids, these mechanical waves cause particles to vibrate, passing energy and sound through the material.
This type of wave has particles oscillating back and forth about their equilibrium positions. Solids are particularly good at transmitting sound because the particles are tightly packed. They transfer energy efficiently from one to another. That's why sound travels faster in solids than in liquids or gases.
When you speak or a bell rings, it creates vibrations that move through the air as mechanical waves. The rate at which sound travels through a solid depends largely on the density and elasticity of that solid. These qualities dictate how quickly particles can transfer the vibrational energy along the wave.
Elasticity and Sound
Elasticity plays a vital role when it comes to sound transmission in materials. It refers to the ability of a material to return to its original shape and size after being deformed. This is critical for how sound waves propagate through a solid.
  • Elastic materials push back against deformations, transferring sound waves quickly.
  • Sound speed in a solid increases with higher elasticity.
In physics, elasticity is often discussed in terms of its coefficient. A high coefficient of elasticity allows solids to transmit sound efficiently. This is why metals and dense materials can carry sound waves faster than less elastic materials.
When sound waves travel through highly elastic materials, the energy loss is minimal, allowing for quicker and more define propagation of sound waves. This is crucial in fields like material science and engineering.
Young's Modulus
Young's Modulus, denoted as \( Y \), is a measure of the stiffness of a solid material. It's a key factor in calculating the speed of sound in solids, with higher values of Young's Modulus indicating greater elasticity.
  • Young's Modulus is defined as the ratio of stress to strain in the material.
  • It provides insight into a material's ability to withstand changes in length when under tension or compression.
The formula for Young's Modulus is \( Y = \frac{F/A}{ riangle L/L} \), where \( F \) is the force applied, \( A \) is the area, \( riangle L \) is the change in length, and \( L \) is the original length.
In terms of sound, materials with a high Young's Modulus are more rigid and allow sound waves to pass through more swiftly. This is why the speed of sound is higher in solids like metals where the Young's Modulus is large. By understanding Young's Modulus, we can predict how different materials might affect sound transmission, aiding in the design of various engineering and technological applications.

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Most popular questions from this chapter

A plane wave \(y=a \sin (b x+c t)\) is incident on a surlace. Fquation of the rellected wave is \(y^{\prime}=a^{\prime} \sin (c t-b x)\). Then which of the following statements are correct? (a) The wave is incident normally on the surface. (b) Rellecting surface is \(y-z\) plane. (c) Medium, in which incident wave is travelling, is denser than the other medium. (d) \(a^{\prime}\) cannot be greater than \(a\).

Statement-1 : 'The frequencies of incident, reflected and refracted beam of monochromatic sound incident from one medium to other medium are same. Statement-2: The incident, rellected and refracted rays are coplanar.

(a) Let \(L \rightarrow\) stationary, \(\nu=580 \mathrm{II} \mathrm{z}\) \(s \rightarrow\) moving with \(40 \mathrm{~km} / \mathrm{hr}\) and aproaching \(w \rightarrow 40 \mathrm{~km} / \mathrm{hr}\) and supporting \(v=1200 \mathrm{~km} / \mathrm{hr}\) So \(v^{\prime}=\left[\frac{v+\omega}{(v+\omega)-v_{s}}\right] v\) \(\Rightarrow \quad v^{\prime}=\frac{1240}{1200} \times 580=\frac{1798}{3}=599.33 \mathrm{IIz}\) (b) At a distance \(1 \mathrm{~km}\) before the train whistles and let the driver heard the echo at time \(l=t^{\prime}\). then at this time, total distance travelled by wave and train \(=2 \mathrm{~km}\) Now time after which first wave reach to the hill in time interval \(t_{1}=\frac{1}{1200} \mathrm{hr}\) after reflecting at \(t=t^{\prime}\), echo will heard, then distance travellcd by it is \(x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200\) and distance travelled by the train is \(x_{2}=\left(t^{\prime}-\frac{1}{1200}\right) 40\) As \(x_{1}+x_{2}=1 \mathrm{~km}\) \(\therefore\left(t^{\prime}-\frac{1}{1200}\right) \times 12040+\left(t^{\prime}-\frac{1}{1200}\right) 40\) \(1240 t^{\prime}=\frac{1}{30} \quad \therefore \quad t^{\prime}=\frac{61}{30 \times 1240}\) and required distance, $$ x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200 $$$=\left(\frac{61}{30 \times 1240}-\frac{1}{1200}\right) \times 1200=\frac{61 \times 1200}{30 \times 1240}-1\( \)=\frac{30}{31} \times 1000 \mathrm{~m}=967.74 \mathrm{~m}\( Frequency of sound heard by the driver, \)v^{\prime \prime}=\left(\frac{(v-\omega)+v_{2}}{v-\omega}\right) v\( \)=\left(\frac{1200-40+40}{1200-40}\right) \times 599.33\( \)\therefore \quad v^{\prime \prime}=\frac{1200}{1160} \times 599.33=620 \mathrm{~Hz}$

Statement-1 : In a simple harmonic oscillation the average potential encrgy in one time period is cqual to the average kinetic energy in this period. Statement-2 : The average kinetic energy for one oscillation is equal to half of total mechanical energy.

Statement-1: Compression and rerefaction involve change in density and pressure . Statement-2: When particles are compressed, density of medium increases and when rarefied density of medium decreases.

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