The frequency of `A` note is `4` times that of `B` note. The energies of two notes are equal. The amplitude of `b` notes as compared to that of `A` note will be

Options

A. double
B. equal
C. four times
D. eight times

Explanation

To solve the problem, we need to analyze the relationship between the frequencies, energies, and amplitudes of the two notes A and B.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Let the frequency of note B be \( f_B \).
- Then the frequency of note A is \( f_A = 4 f_B \) (since the frequency of A is 4 times that of B).
- The energies of the two notes are equal, i.e., \( E_A = E_B \).

2. **Energy of Oscillations:**
- The energy of a wave is given by the formula:
\[ E \propto A^2 f^2 \]
where \( A \) is the amplitude and \( f \) is the frequency of the wave.
- Therefore, for notes A and B, we can write:
\[ E_A \propto A_A^2 f_A^2 \]
\[ E_B \propto A_B^2 f_B^2 \]

3. **Setting Energies Equal:**
- Since \( E_A = E_B \), we have:
\[ A_A^2 f_A^2 = A_B^2 f_B^2 \]

4. **Substituting Frequencies:**
- Substitute \( f_A = 4 f_B \) into the equation:
\[ A_A^2 (4 f_B)^2 = A_B^2 f_B^2 \]
- This simplifies to:
\[ A_A^2 \cdot 16 f_B^2 = A_B^2 f_B^2 \]

5. **Canceling \( f_B^2 \):**
- Since \( f_B^2 \) is common on both sides, we can cancel it out:
\[ 16 A_A^2 = A_B^2 \]

6. **Finding the Relationship Between Amplitudes:**
- Taking the square root of both sides gives:
\[ A_B = 4 A_A \]

7. **Conclusion:**
- The amplitude of note B is 4 times the amplitude of note A. Therefore, the amplitude of B compared to A is:
\[ \text{Amplitude of B} = 4 \times \text{Amplitude of A} \]

### Final Answer:
The amplitude of B notes as compared to that of A notes will be **4 times**.

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