minimum frequency of audible sound is 20 Hz and maximum frequency is 20 , 000 Hz. If we compare the sound levels of these two frequencies of same amplitudes, their difference will be

Options

A. 30 dB
B. 60 dB
C. 90 dB
D. 120 dB

Answer

To find the difference in sound levels between two frequencies of the same amplitude, we can follow these steps:

### Step 1: Understand the relationship between intensity and frequency
The intensity of sound is proportional to the square of the frequency. This can be expressed as:
\[ I \propto f^2 \]
Where \( I \) is the intensity and \( f \) is the frequency.

### Step 2: Set up the intensity ratio for the two frequencies
Let \( f_1 = 20 \, \text{Hz} \) (minimum frequency) and \( f_2 = 20,000 \, \text{Hz} \) (maximum frequency). The ratio of intensities can be expressed as:
\[ \frac{I_1}{I_2} = \left(\frac{f_1}{f_2}\right)^2 \]

### Step 3: Substitute the values of frequencies into the intensity ratio
Substituting the values of \( f_1 \) and \( f_2 \):
\[ \frac{I_1}{I_2} = \left(\frac{20}{20,000}\right)^2 \]
\[ = \left(\frac{1}{1000}\right)^2 \]
\[ = \frac{1}{1,000,000} \]

### Step 4: Find the ratio of intensities
This means:
\[ \frac{I_2}{I_1} = 1,000,000 = 10^6 \]

### Step 5: Calculate the difference in sound levels
The sound level in decibels (dB) is given by:
\[ L = 10 \log \left(\frac{I}{I_0}\right) \]
To find the difference in sound levels \( \Delta L \) between the two frequencies, we can use:
\[ \Delta L = 10 \log \left(\frac{I_2}{I_1}\right) \]
Substituting the intensity ratio:
\[ \Delta L = 10 \log (10^6) \]
\[ = 10 \times 6 \]
\[ = 60 \, \text{dB} \]

### Conclusion
The difference in sound levels between the two frequencies of the same amplitude is **60 dB**.

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