A progressive wave of frequency 500 Hz is travelling with a speed of 350 m/s. A compressional maximum appears at a place at a given instant. The minimum time interval after which of refraction maximum occurs at the same place is

Options

A. `(1)/(250)s`
B. `(1)/(500)s`
C. `(1)/(1000)s`
D. `(1)/(350)s`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript:

### Step 1: Identify the given values
- Frequency (f) = 500 Hz
- Speed of the wave (v) = 350 m/s

### Step 2: Calculate the time period (T) of the wave
The time period (T) is the reciprocal of the frequency (f):
\[ T = \frac{1}{f} \]
Substituting the value of frequency:
\[ T = \frac{1}{500} \text{ seconds} \]

### Step 3: Understand the relationship between compressional maximum and rarefaction maximum
In one complete cycle (one time period), both a compressional maximum and a rarefaction maximum occur. This means:
- In one time period (T), there is one compressional maximum and one rarefaction maximum.

### Step 4: Calculate the time interval for the next rarefaction maximum
Since both a compressional maximum and a rarefaction maximum occur in one complete cycle, the time interval after which a rarefaction maximum occurs after a compressional maximum is half of the time period:
\[ \text{Time interval} = \frac{T}{2} \]
Substituting the value of T:
\[ \text{Time interval} = \frac{1/500}{2} = \frac{1}{1000} \text{ seconds} \]

### Step 5: Conclusion
Thus, the minimum time interval after which a rarefaction maximum occurs at the same place is:
\[ \frac{1}{1000} \text{ seconds} \]
The correct answer is option C: \( \frac{1}{1000} \) seconds.

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