A source of sound gives 5 beats when sounded with another source of frequency 100 Hz. The second harmonic of the source together with a source of frequency 200 Hz gives 10 beats `s^(-1)`. What is the frequency of the source?

Options

A. 105 Hz
B. 205 Hz
C. 90 Hz
D. 100 Hz

Answer

To solve the problem, we will analyze the information given step by step.

### Step 1: Understanding Beats
When two sound sources of different frequencies are sounded together, the number of beats per second is equal to the absolute difference between their frequencies.

Given:
- The first source (unknown frequency) gives 5 beats with a source of frequency 100 Hz.

Let the frequency of the unknown source be \( f \). According to the beat frequency formula:
\[ |f - 100| = 5 \]
This means:
\[ f - 100 = 5 \quad \text{or} \quad f - 100 = -5 \]
From this, we can derive two possible values for \( f \):
1. \( f = 105 \, \text{Hz} \)
2. \( f = 95 \, \text{Hz} \)

### Step 2: Analyzing the Second Condition
The second part of the problem states that the second harmonic of the source, together with a source of frequency 200 Hz, gives 10 beats per second.

The second harmonic of the source frequency \( f \) is given by:
\[ 2f \]
Now, using the beat frequency formula again:
\[ |2f - 200| = 10 \]
This gives us two equations:
1. \( 2f - 200 = 10 \)
2. \( 2f - 200 = -10 \)

### Step 3: Solving the Equations
**For the first equation:**
\[ 2f - 200 = 10 \implies 2f = 210 \implies f = 105 \, \text{Hz} \]

**For the second equation:**
\[ 2f - 200 = -10 \implies 2f = 190 \implies f = 95 \, \text{Hz} \]

### Step 4: Conclusion
From the two conditions, we have derived two possible frequencies for the source:
- \( f = 105 \, \text{Hz} \)
- \( f = 95 \, \text{Hz} \)

Both frequencies satisfy the conditions given in the problem. Therefore, the frequency of the source can be either 105 Hz or 95 Hz.

### Final Answer
The frequency of the source is either **105 Hz or 95 Hz**.

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