Question

What must be the stress \((F/A)\) in a stretched wire of a material whose Young's modulus is \(Y\) for the speed of longitudinal waves to equal 30 times the speed of transverse waves?

Short answer

The stress must be \( \frac{Y}{900} \).

Step-by-step solution

Understanding Wave Speeds

The speed of longitudinal waves in a material is given by \( v_l = \sqrt{\frac{Y}{\rho}} \), where \( Y \) is Young's modulus and \( \rho \) is the density of the material. The speed of transverse waves \( v_t \) in a wire is given by \( v_t = \sqrt{\frac{F}{\mu}} \), where \( F \) is the tension in the wire and \( \mu \) (linear density) is \( \rho \cdot A \) for a wire of cross-sectional area \( A \).

Expressing the Given Condition

According to the problem, the speed of longitudinal waves is 30 times the speed of transverse waves, hence \( v_l = 30 \cdot v_t \). We can substitute the expressions for \( v_l \) and \( v_t \), leading to: \[ \sqrt{\frac{Y}{\rho}} = 30 \cdot \sqrt{\frac{F}{\mu}} \] .

Solving for the Tension-to-Area Ratio

Equate the squares of both sides of the equation obtained: \( \frac{Y}{\rho} = 30^2 \cdot \frac{F}{\rho \cdot A} \). Note that \( \mu = \rho \cdot A \).

Cancelling and Rearranging Terms

Simplify the equation: \( \frac{Y}{\rho} = 900 \cdot \frac{F}{\rho \cdot A} \). Cancel \( \rho \) from both sides to get \( Y = 900 \cdot \frac{F}{A} \).

Finding the Stress

Rearrange the equation to solve for the stress \( \frac{F}{A} \): \( \frac{F}{A} = \frac{Y}{900} \).

Conclusion

The stress required in the wire for the condition given in the problem is \( \frac{Y}{900} \).

Key concepts

Wave Speed

Wave speed is an essential concept in physics that tells us how quickly a wave travels through a medium. In the context of this exercise, we are dealing with two types of wave speeds: longitudinal and transverse.
For longitudinal waves in a material, the speed, denoted as \( v_l \), is derived from the equation:

• \( v_l = \sqrt{\frac{Y}{\rho}} \)

Here, \( Y \) is Young's modulus, a measure of the stiffness of the material, and \( \rho \) represents the density of the material.
On the other hand, the speed of transverse waves, \( v_t \), in a wire is expressed as:

• \( v_t = \sqrt{\frac{F}{\mu}} \)

In this formula, \( F \) is the tension force, and \( \mu \) is the linear density of the wire, defined as the product of the material's density and the cross-sectional area \( A \).
Understanding these wave speeds is crucial for solving problems related to wave motion.

Stress in a Wire

Stress in a wire is an important concept when dealing with the elasticity of materials. Stress is the force \( F \) applied per unit area \( A \), represented by the equation:

• Stress \( = \frac{F}{A} \)

In this exercise, we are asked to find the stress that will result in the speed of longitudinal waves in a wire being thirty times that of transverse waves.
To solve this, we use the equations for wave speeds and set up the equation:

• \( \sqrt{\frac{Y}{\rho}} = 30 \cdot \sqrt{\frac{F}{\mu}} \)

By rearranging and simplifying, we find:

• Stress \( = \frac{Y}{900} \)

This tells us the exact value of stress needed in the wire when the given condition is satisfied.

Longitudinal Waves

Longitudinal waves are waves where the particles in the material move in the same direction as the wave. In the context of a wire, these waves occur when a material is compressed and stretched along the axis of the wave.
The speed at which these waves move through a medium depends on the material's properties, specifically its Young's modulus \( Y \) and density \( \rho \). The formula to compute this speed is:

• \( v_l = \sqrt{\frac{Y}{\rho}} \)

This type of wave speed is typically higher than transverse wave speeds when considering solid materials like metals. Understanding this allows us to comprehend how stress and material properties influence wave motion.

Transverse Waves

Transverse waves, unlike longitudinal waves, involve particle motion perpendicular to the direction of wave propagation. In a wire, such waves occur when it undergoes displacement sideways under tension.
The speed of transverse waves \( v_t \) is determined by the tension in the wire \( F \) and its linear density \( \mu \). The expression for the speed is:

• \( v_t = \sqrt{\frac{F}{\mu}} \)

Linear density \( \mu \) is calculated as the product of the wire's cross-sectional area \( A \) and its material density \( \rho \).
These waves are slower than longitudinal waves in the same material due to the nature of their particle motion. This difference in speed helps us understand why the stress in the wire needs to be precisely calculated to achieve the exercise's requirements.