Question

If the amplitude of sound is doubled and the frequency reduced to one fourth, the intensity of sound at the same point will be

• A. increasing by a factor of 2
• B. decreasing by a factor of 2
• C. decreasing by a factor of 1/4
• D. unchanged

Answer 1

The Correct Option is C

To determine the effect on the intensity of sound when the amplitude is doubled and the frequency is reduced to one fourth, we must understand how intensity is related to amplitude and frequency.

The intensity of a sound wave, \(I\), is proportional to the square of its amplitude, \(A\), and the square of its frequency, \(f\). Mathematically, this is expressed as:

I \propto A^2 f^2

Let's analyze the changes step-by-step:

1. Doubling the amplitude:
   • If the amplitude \(A\) is doubled, the new amplitude becomes \(2A\).
   • The new intensity due to change in amplitude is proportional to \((2A)^2 = 4A^2\).
2. Reducing the frequency to one fourth:
   • If the frequency \(f\) is reduced to one fourth, the new frequency becomes \(\frac{f}{4}\).
   • The new intensity due to change in frequency is proportional to \(\left(\frac{f}{4}\right)^2 = \frac{f^2}{16}\).

Combining both changes, the new intensity is proportional to:

I_{\text{new}} \propto (2A)^2 \left(\frac{f}{4}\right)^2 = 4A^2 \times \frac{f^2}{16} = \frac{A^2 f^2}{4}

Compared to the original intensity, the new intensity is:

I_{\text{new}} = \frac{1}{4} I_{\text{original}}

Thus, the intensity of sound decreases by a factor of \(\frac{1}{4}\), which matches with the given option: decreasing by a factor of 1/4.