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\(\delta=\frac{m}{L}=\frac{32}{4} \times 10^{2}=\frac{4}{5} \times 10^{2} \mathrm{~kg} / \mathrm{m}\) Now, According to question, \(v=v_{o}=220 \mathrm{II} z\) \(\therefore \frac{v}{2 L}=220 \mathrm{~Hz} \quad[\quad L=40 \mathrm{~cm}\) \(m=3.2 \mathrm{gm}\) \(\sqrt{\frac{T}{\delta}} \times \frac{1}{2 L}=220 \mathrm{II} z\) 1. \(\sqrt{\frac{T}{\rho}}=220 \times 2 L\) \(\therefore \quad \sqrt{T}=\sqrt{\rho} \times 220 \times 2 L$$$ \begin{aligned} &\therefore \quad T=\rho \times(220 \times 2 L)^{2}=\frac{4}{5} \times 10^{-2}(22 \times 2 \times 4)^{2} \\ &=\frac{4}{5} \times 10^{-2} \times 30976=2478.8 \times 10^{-2} \\ &\text { Now suress }=\frac{F}{\Lambda}=\frac{T}{\Lambda}=\frac{2478.8 \times 10^{-2}}{\left(10^{-3}\right)^{2}} \\ &=2478.8 \times 10^{4}=\mathrm{N} / \mathrm{m}^{2} \end{aligned} $$ and strain $$ =\frac{0.05}{40} \times \frac{10^{2}}{10^{2}}=\frac{5}{40} \times 10^{-2}=\frac{1}{8} \times 10^{-2} $$ \)\therefore$ (Youngs modulus) $$ \begin{aligned} Y &=\frac{\text { stress }}{\text { strain }}=\frac{2478.8 \times 10^{4}}{10^{2} / 8} \\ &=1.98 \times 101^{\prime} \mathrm{N} / \mathrm{m}^{2} \end{aligned} $$

Step by step solution

01

Calculate Linear Mass Density

The linear mass density \( \delta \) is calculated using the formula \( \delta = \frac{m}{L} \). Given that mass \( m = 32 \) grams and length \( L = 4 \) m, first convert \( m \) to kilograms: \( 32 \times 10^{-3} \) kg. Then, \( \delta = \frac{32 \times 10^{-3}}{4} = \frac{32}{4} \times 10^{-3} = 8 \times 10^{-3} \). Convert to proper units as \( 8 \times 10^{-2} \) kg/m.

02

Use Given Frequency Formula

Using \( v = v_0 = 220 \) Hz and \( L = 40 \) cm \( = 0.4 \) m, the formula is \( \frac{v}{2L} = 220 \). This implies \( v = 2L \times 220 \), showing the relationship between the speed of the wave, frequency, and length.

03

Solve for Tension

From the wave speed formula \( \sqrt{\frac{T}{\delta}} = \frac{v}{2L} = 220 \), substitute for \( v = 220 \times 2L \). Rearrange to \( T = \rho \times (220 \times 2L)^2 \) where \( \rho = \delta = 8 \times 10^{-2} \). Calculate tension, \( T = 8 \times 10^{-2} \times (220 \times 0.8)^2 = 2478.8 \times 10^{-2} \).

04

Calculate Stress

Stress is calculated using \( \text{Stress} = \frac{T}{\Lambda} \) where, \( \Lambda = 10^{-6} \). Thus, Stress = \( \frac{2478.8 \times 10^{-2}}{10^{-6}} = 2478.8 \times 10^{4} \text{ N/m}^2 \).

05

Calculate Strain

Strain is defined as \( \frac{\Delta L}{L} \). With given \( \Delta L = 0.05 \) m and \( L = 0.4 \) m: \( \text{Strain} = \frac{0.05}{0.4} = \frac{1}{8} = 0.125 \).

06

Calculate Young's Modulus

Young's modulus \( Y \) is calculated as \( \frac{\text{Stress}}{\text{Strain}} \). With \( \text{Stress} = 2478.8 \times 10^{4} \text{ N/m}^2 \) and \( \text{Strain} = 0.125 \), Young's modulus \( Y = \frac{2478.8 \times 10^{4}}{0.125} = 1.98 \times 10^{6} \text{ N/m}^2 \).

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Key Concepts

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

Linear Mass Density

When analyzing waves, particularly in strings, understanding linear mass density is crucial. Linear mass density, represented by \( \delta \), essentially tells us how much mass exists per unit length of a string. This is calculated by taking the mass \( m \) of the string and dividing it by the length \( L \) of the string: \( \delta = \frac{m}{L} \).
In the exercise, the mass is given as 32 grams, which we convert to kilograms because it's a standard unit for such calculations—resulting in \( 32 \times 10^{-3} \) kg. The length is 4 meters. By plugging these values into the formula, we're able to determine the linear mass density: \( 8 \times 10^{-2} \) kg/m.

This value is essential, as it helps in determining other wave-related properties like tension and speed within the string.

Wave Speed and Frequency Relationship

The relationship between wave speed and frequency in a string is a fundamental concept in understanding wave mechanics. In this scenario, the speed of the wave (\( v \)) and its frequency (\( 220 \) Hz, as given) relate through the formula \( \frac{v}{2L} = 220 \). Here, \( L \) is the length of the string, expressed in meters (0.4 m).
From this relationship, we can extract the formula \( v = 2L \times 220 \). This equation shows how the speed of the wave is affected by both its frequency and the length of the medium it travels through. By knowing any two of these values, you can calculate the third, which is especially useful in practical scenarios like tuning musical instruments or designing systems that involve wave propagation.

Stress and Strain Calculations

Stress and strain are vital concepts when examining materials under tension, and they are particularly important in calculating Young's modulus. Stress is defined as the force exerted per unit area, with its formula \( \text{Stress} = \frac{T}{\Lambda} \). Here \( T \) represents tension and \( \Lambda \) is a cross-sectional area. In the exercise, Stress comes out as \( 2478.8 \times 10^{4} \text{ N/m}^2 \).
Strain, on the other hand, measures the deformation of the string, essentially the change in length (\( \Delta L \)) over its original length \( L \). It is calculated by \( \text{Strain} = \frac{\Delta L}{L} \). In the provided solution, Strain is \( \frac{1}{8} = 0.125 \).
Finally, Young's modulus, denoted by \( Y \), is a measure of the stiffness of a material. It is calculated using the formula \( Y = \frac{\text{Stress}}{\text{Strain}} \). With the given values, it results in \( 1.98 \times 10^{6} \text{ N/m}^2 \), a crucial attribute for understanding how the string will behave under various forces.

Tension in a String

Tension is a key factor that influences the behavior of waves in a string. It denotes the force applied along the string, affecting its vibrations and wave transmission properties. For a string to convey waves with consistent speed and frequency, tension must be properly calculated.
The exercise employs the formula \( \sqrt{\frac{T}{\delta}} = \frac{v}{2L} \), indicating that the wave speed depends on both tension and linear mass density. Solving for tension \( T \), we get \( T = \delta \times (220 \times 0.8)^2 = 2478.8 \times 10^{-2} \), indicating how tension is derived from the given wave speed relationship and linear mass density.
Understanding tension helps us adjust musical instruments or predict the physical behavior of strings in various applications, from engineering to acoustic studies.