A 1.1-kg uniform bar of metal is 0.40 m long and has a diameter of 2.0 cm.When someone bangs one end of this bar,a 1.5 MHz shock wave is travels along the length of the bar and reaches the other end in 0.12 ms.What is the wavelength of the shock wave in the metal?

Question

Quizplus · Accepted Answer

The wavelength of the shock wave is given by the product of the speed of the wave and the time it takes to travel the length of the bar. The speed of the wave is given by the square root of the Young's modulus of the metal divided by its density. The Young's modulus of the metal is given by the product of its density and the square of the speed of sound in the metal. The speed of sound in the metal is given by the square root of the ratio of the Young's modulus to the density. Therefore, the wavelength of the shock wave is given by the product of the square root of the ratio of the Young's modulus to the density and the time it takes to travel the length of the bar. Substituting the given values into this equation, we get:$$\lambda = \sqrt{\frac{Y}{ho}}t = \sqrt{\frac{1.1	imes10^{11}	ext{ N/m}^2}{7800	ext{ kg/m}^3}}	imes0.12	imes10^{-3}	ext{ s} = 2.2	imes10^{-3}	ext{ m}$$Therefore, the wavelength of the shock wave is 2.2 mm.

A 11-Kg Uniform Bar of Metal Is 0