Question

If percentage increase in Young's modulus \(Y\) is \(1%\), percentage increase in density of material is \(0.5%\) and longitudinal wave traveling in a metallic bar has wave velocity of \(400\,\text{m/s}\), then find final velocity of the wave.

• A. \(398\,\text{m/s}\)
• B. 355
• C. 401
• D. 402

Hint:

For wave speed relations involving square roots, always remember: \(\displaystyle \frac{\Delta v}{v} = \frac{1}{2}\frac{\Delta X}{X}\). Apply signs carefully when ratios are involved.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between the wave velocity in a material, its Young's modulus, and its density. The velocity of a longitudinal wave \( v \) in a material is given by the formula:

v = \sqrt{\frac{Y}{\rho}}

where \( Y \) is the Young's modulus and \( \rho \) is the density of the material.

According to the problem statement, there is a 1% increase in Young's modulus and a 0.5% increase in the density of the material. First, let's express these increases mathematically:

• If the initial Young's modulus is \( Y \), then the new Young's modulus becomes \( Y' = Y(1 + 0.01) \).
• If the initial density is \( \rho \), then the new density becomes \( \rho' = \rho(1 + 0.005) \).

We are given the initial wave velocity \( v = 400\,\text{m/s} \). The final velocity \( v' \) after the changes can be expressed as:

v' = \sqrt{\frac{Y'}{\rho'}} = \sqrt{\frac{Y(1 + 0.01)}{\rho(1 + 0.005)}}

Simplifying the equation gives:

v' = v \times \sqrt{\frac{1.01}{1.005}}

Now, we calculate \( v' \):

v' = 400 \times \sqrt{\frac{1.01}{1.005}} \approx 400 \times 1.00249 \approx 400.996 \approx 401 \, \text{m/s}

Thus, the final velocity of the wave is approximately \( 401 \, \text{m/s} \).

Therefore, the correct answer is 401.