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Waves - The Speed of a Travelling Wave

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The speed of a mechanical wave is determined by the inertial and elastic properties of the medium through which it propagates.

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For a transverse wave on a stretched string, the speed vv depends on the tension TT (elastic property) and the linear mass density μ\mu (inertial property).

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Linear mass density μ\mu is defined as mass per unit length of the string, given by μ=mL\mu = \frac{m}{L}.

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In a longitudinal wave (like sound), the speed depends on the Bulk modulus BB and the density ρ\rho of the medium.

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Newton's formula for the speed of sound in an ideal gas assumed the process is isothermal, leading to v=Pρv = \sqrt{\frac{P}{\rho}}.

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Laplace corrected Newton's formula by noting that sound propagation is an adiabatic process. The corrected formula is v=γPρv = \sqrt{\frac{\gamma P}{\rho}}, where γ\gamma is the ratio of specific heats (Cp/CvC_p/C_v).

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The general relation between speed vv, frequency ν\nu, and wavelength λ\lambda for any periodic wave is v=νλv = \nu \lambda.

📐Formulae

v=λT=νλv = \frac{\lambda}{T} = \nu \lambda

v=Tμv = \sqrt{\frac{T}{\mu}}

v=Bρ (for fluids)v = \sqrt{\frac{B}{\rho}} \text{ (for fluids)}

v=Yρ (for solid rods)v = \sqrt{\frac{Y}{\rho}} \text{ (for solid rods)}

v=γPρ=γRTM (Laplace’s correction)v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}} \text{ (Laplace's correction)}

💡Examples

Problem 1:

A steel wire 0.72 m0.72\text{ m} long has a mass of 5.0×10−3 kg5.0 \times 10^{-3}\text{ kg}. If the wire is under a tension of 60 N60\text{ N}, what is the speed of transverse waves on the wire?

Solution:

Given: Length L=0.72 mL = 0.72\text{ m}, Mass m=5.0×10−3 kgm = 5.0 \times 10^{-3}\text{ kg}, Tension T=60 NT = 60\text{ N}. First, calculate linear mass density μ\mu: μ=mL=5.0×10−30.72≈6.94×10−3 kg m−1\mu = \frac{m}{L} = \frac{5.0 \times 10^{-3}}{0.72} \approx 6.94 \times 10^{-3}\text{ kg m}^{-1} Now, use the wave speed formula: v=Tμ=606.94×10−3=8645.5≈93 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{60}{6.94 \times 10^{-3}}} = \sqrt{8645.5} \approx 93\text{ m/s}

Explanation:

The speed is calculated by finding the ratio of tension to mass per unit length and then taking the square root. Higher tension leads to higher wave speed.

Problem 2:

Estimate the speed of sound in air at standard temperature and pressure (STP) using Laplace's correction. (Given: γ=1.4\gamma = 1.4, P=1.013×105 PaP = 1.013 \times 10^5\text{ Pa}, ρ=1.29 kg/m3\rho = 1.29\text{ kg/m}^3)

Solution:

Using Laplace's formula: v=γPρv = \sqrt{\frac{\gamma P}{\rho}} Substitute the values: v=1.4×1.013×1051.29v = \sqrt{\frac{1.4 \times 1.013 \times 10^5}{1.29}} v=1418201.29=109937.98≈331.57 m/sv = \sqrt{\frac{141820}{1.29}} = \sqrt{109937.98} \approx 331.57\text{ m/s}

Explanation:

Laplace's correction incorporates the adiabatic index γ\gamma because the compressions and rarefactions in a sound wave happen too rapidly for heat exchange with the surroundings.