Question

The speed of a longitudinal wave in a metallic bar is \(400\,\text{m/s}\). If the density and Young’s modulus of the bar material increase by \(0.5%\) and \(1%\) respectively, then the speed of the wave is changed approximately to _______ m/s.

• A. \(399\)
• B. \(398\)
• C. \(402\)
• D. \(401\)

Hint:

For quantities under square root, remember that relative error is halved.

Answer 1

The Correct Option is D

To determine the new speed of the longitudinal wave in the metallic bar, we use the formula for the speed of a wave in a solid medium given by:

\(v = \sqrt{\frac{E}{\rho}}\)

where:

• \(v\) is the speed of sound (wave speed).
• \(E\) is the Young's modulus of the material.
• \(\rho\) is the density of the material.

 

Initially, the speed of the wave \(v_1\) is given as \(400\,\text{m/s}\).

When the density increases by 0.5%, the new density will be: \(\rho' = \rho + 0.005\rho = 1.005\rho\)

Similarly, when Young's modulus increases by 1%, the new modulus will be: \(E' = E + 0.01E = 1.01E\)

The new speed of the wave \(v_2\) is calculated as:

\(v_2 = \sqrt{\frac{E'}{\rho'}}\)

Substituting the changes in \(E\) and \(\rho\), we have:

\(v_2 = \sqrt{\frac{1.01E}{1.005\rho}}\)

\(= \sqrt{\frac{1.01}{1.005}} \cdot \sqrt{\frac{E}{\rho}}\)

Since \(\sqrt{\frac{E}{\rho}} = v_1 = 400 \, \text{m/s}\), substitute in:

\(v_2 = \sqrt{\frac{1.01}{1.005}} \times 400\)

Using a calculator to find \(\sqrt{\frac{1.01}{1.005}} \approx 1.0025\),

\(v_2 = 1.0025 \times 400 \approx 401 \, \text{m/s}\)

Therefore, the new speed of the wave in the metallic bar is approximately \(401\,\text{m/s}\).

This matches the provided correct answer option: \(401\).