The number of compressions and rarefactions which pass an observer in 5 minutes, if the velocity of sound wave is 330 m/s and the wavelength of the wave is 30 cm, is

Options

A. 3300
B. `3.3xx10^(4)`
C. `3.3xx10^(5)`
D. `3.3xx10^(2)`

Answer

To solve the problem of finding the number of compressions and rarefactions that pass an observer in 5 minutes, we can follow these steps:

### Step 1: Convert Time from Minutes to Seconds
Given that the time is 5 minutes, we need to convert this into seconds.

\[ \text{Time in seconds} = 5 \text{ minutes} \times 60 \text{ seconds/minute} = 300 \text{ seconds} \]

### Step 2: Write Down the Given Values
- Velocity of sound, \( v = 330 \text{ m/s} \)
- Wavelength, \( \lambda = 30 \text{ cm} = 0.3 \text{ m} \)

### Step 3: Calculate Frequency
We can use the formula that relates the speed of sound, wavelength, and frequency:

\[ f = \frac{v}{\lambda} \]

Substituting the values:

\[ f = \frac{330 \text{ m/s}}{0.3 \text{ m}} = 1100 \text{ Hz} \]

### Step 4: Calculate the Number of Compressions and Rarefactions in 300 Seconds
Since the frequency \( f \) tells us the number of compressions (or rarefactions) that pass an observer in one second, we can calculate the total number in 300 seconds:

\[ \text{Number of compressions and rarefactions} = f \times \text{Time} \]

Substituting the values:

\[ \text{Number} = 1100 \text{ Hz} \times 300 \text{ seconds} = 330000 \]

### Step 5: Express the Result in Scientific Notation
We can express 330000 in scientific notation:

\[ 330000 = 3.3 \times 10^5 \]

### Final Answer
The number of compressions and rarefactions that pass an observer in 5 minutes is \( 3.3 \times 10^5 \).

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