The frequency of a wave is reduced to one quarter and its amplitude is made twice. The intensity ofthe wave:

Options

A. Increases by a factor of 2
B. Decreases by a factor of 4
C. Decreasesbyafectorof2
D. Remain unchanged

Explanation

To solve the problem, we will use the relationship between intensity, amplitude, and frequency of a wave. The intensity \( I \) of a wave is given by the formula:

\[ I \propto A^2 f^2 \]

where:
- \( I \) is the intensity,
- \( A \) is the amplitude,
- \( f \) is the frequency.

### Step 1: Define the initial intensity
Let the initial amplitude be \( A_1 \) and the initial frequency be \( f_1 \). The initial intensity \( I_1 \) can be expressed as:

\[ I_1 = k A_1^2 f_1^2 \]

where \( k \) is a proportionality constant.

### Step 2: Define the new amplitude and frequency
According to the problem:
- The frequency is reduced to one quarter:
\[ f_2 = \frac{f_1}{4} \]
- The amplitude is made twice:
\[ A_2 = 2 A_1 \]

### Step 3: Calculate the new intensity
Now, we can express the new intensity \( I_2 \) using the new amplitude and frequency:

\[ I_2 = k A_2^2 f_2^2 \]

Substituting the values of \( A_2 \) and \( f_2 \):

\[ I_2 = k (2 A_1)^2 \left(\frac{f_1}{4}\right)^2 \]

### Step 4: Simplify the expression
Now, simplify the expression for \( I_2 \):

\[ I_2 = k (4 A_1^2) \left(\frac{f_1^2}{16}\right) \]
\[ I_2 = k A_1^2 f_1^2 \cdot \frac{4}{16} \]
\[ I_2 = k A_1^2 f_1^2 \cdot \frac{1}{4} \]
\[ I_2 = \frac{1}{4} I_1 \]

### Step 5: Conclusion
This shows that the new intensity \( I_2 \) is one quarter of the initial intensity \( I_1 \):

\[ I_2 = \frac{I_1}{4} \]

Thus, the intensity of the wave is reduced by a factor of 4.

### Final Answer
The intensity of the wave is reduced to one quarter of its original intensity.

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