A source `S` of acoustic wave of the frequency `v_0=1700Hz` and a receiver `R` are located at the same point. At the instant `t=0`, the source start from rest to move away from the receiver with a constant acceleration `omega`. The velocity of sound in air is `v=340(m)/(s)`.
If `omega=10(m)/(s^2)` for 10s and then `omega=0` for `tgt10s`, the apparent frequency recorded by the receiver at `t=15s`

Options

A. `1700Hz`
B. `1313Hz`
C. `850Hz`
D. `1.23Hz`

Explanation

To solve the problem step by step, we will follow the principles of kinematics and the Doppler effect.

### Step 1: Determine the velocity of the source at \( t = 10 \) seconds
The source starts from rest and accelerates with a constant acceleration \( \omega = 10 \, \text{m/s}^2 \) for \( 10 \) seconds. We can use the first equation of motion:

\[ v = u + at \]

where:
- \( u = 0 \) (initial velocity),
- \( a = 10 \, \text{m/s}^2 \) (acceleration),
- \( t = 10 \, \text{s} \).

Substituting the values:

\[ v = 0 + (10 \, \text{m/s}^2)(10 \, \text{s}) = 100 \, \text{m/s} \]

### Step 2: Understand the motion after \( t = 10 \) seconds
After \( t = 10 \) seconds, the acceleration becomes \( 0 \), meaning the source will continue to move at a constant velocity of \( 100 \, \text{m/s} \) for the next \( 5 \) seconds (from \( t = 10 \) to \( t = 15 \) seconds).

### Step 3: Apply the Doppler Effect formula
The apparent frequency \( f' \) can be calculated using the Doppler effect formula:

\[ f' = f \frac{v - v_0}{v - v_s} \]

where:
- \( f = 1700 \, \text{Hz} \) (frequency of the source),
- \( v = 340 \, \text{m/s} \) (velocity of sound in air),
- \( v_0 = 0 \, \text{m/s} \) (velocity of the observer, since the receiver is stationary),
- \( v_s = 100 \, \text{m/s} \) (velocity of the source at \( t = 15 \) seconds).

### Step 4: Substitute the values into the formula
Substituting the known values into the Doppler effect formula:

\[ f' = 1700 \, \text{Hz} \cdot \frac{340 - 0}{340 - 100} \]

Calculating the denominator:

\[ 340 - 100 = 240 \]

Now substituting back into the formula:

\[ f' = 1700 \, \text{Hz} \cdot \frac{340}{240} \]

### Step 5: Simplify the expression
Calculating the fraction:

\[ \frac{340}{240} = \frac{17}{12} \approx 1.4167 \]

Now calculate \( f' \):

\[ f' = 1700 \, \text{Hz} \cdot 1.4167 \approx 2413.39 \, \text{Hz} \]

### Step 6: Final Calculation
Calculating the final value:

\[ f' \approx 2413.39 \, \text{Hz} \]

### Conclusion
The apparent frequency recorded by the receiver at \( t = 15 \, \text{s} \) is approximately \( 2413.39 \, \text{Hz} \).

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