A source frequency f gives 5 beats when sounded with a frequency 200Hz. The second harmonic of same source gives 10 beats when sounded with a source of frequency 420Hz. The value of f is

Options

A. 205Hz
B. 195 Hz
C. 200 Hz
D. 210 Hz

Answer

To solve the problem, we need to find the value of the frequency \( f \) of a source that produces beats with two different frequencies under two different conditions.

### Step-by-Step Solution:

1. **Understanding Beat Frequency**:
The beat frequency is given by the formula:
\[ f_{\text{beat}} = |f_1 - f_2| \]
where \( f_1 \) and \( f_2 \) are the frequencies of the two sources.

2. **First Condition**:
We know that the source frequency \( f \) gives 5 beats when sounded with a frequency of 200 Hz. Thus, we can write:
\[ |f - 200| = 5 \]
This leads to two possible equations:
\[ f - 200 = 5 \quad \text{or} \quad f - 200 = -5 \]
Solving these gives:
- From \( f - 200 = 5 \):
\[ f = 205 \, \text{Hz} \]
- From \( f - 200 = -5 \):
\[ f = 195 \, \text{Hz} \]

3. **Second Condition**:
The second harmonic of the same source gives 10 beats when sounded with a frequency of 420 Hz. The second harmonic \( f_2 \) is given by:
\[ f_2 = 2f \]
Thus, we have:
\[ |2f - 420| = 10 \]
This leads to two possible equations:
\[ 2f - 420 = 10 \quad \text{or} \quad 2f - 420 = -10 \]
Solving these gives:
- From \( 2f - 420 = 10 \):
\[ 2f = 430 \implies f = 215 \, \text{Hz} \]
- From \( 2f - 420 = -10 \):
\[ 2f = 410 \implies f = 205 \, \text{Hz} \]

4. **Finding Common Value**:
Now we have two possible values for \( f \) from the first condition: \( 205 \, \text{Hz} \) and \( 195 \, \text{Hz} \). From the second condition, we have \( 215 \, \text{Hz} \) and \( 205 \, \text{Hz} \). The common value from both conditions is:
\[ f = 205 \, \text{Hz} \]

### Final Answer:
The value of \( f \) is \( 205 \, \text{Hz} \).

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