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)`.
Q. If `omega=10(m)/(s^2)`, the apparent frequency that will be recorded by the stationary receiver at `t=10s` will be

Options

A. `1700Hz`
B. `1.35Hz`
C. `850Hz`
D. `1.27Hz`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the motion of the source
The source `S` starts from rest and moves away from the receiver `R` with a constant acceleration `ω = 10 m/s²`. We need to find the position of the source after `t = 10 seconds`.

**Hint:** Use the kinematic equation for uniformly accelerated motion:
\[ d = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity), \( a = \omega \), and \( t = 10 \, \text{s} \).

### Step 2: Calculate the distance traveled by the source
Using the kinematic equation:
\[ d = 0 \cdot 10 + \frac{1}{2} \cdot 10 \cdot (10)^2 \]
\[ d = \frac{1}{2} \cdot 10 \cdot 100 = 500 \, \text{m} \]

**Hint:** Remember that the distance traveled is directly proportional to the square of the time when starting from rest.

### Step 3: Determine the velocity of the source at t = 10 seconds
The velocity of the source at time `t` can be calculated using:
\[ v = u + at \]
\[ v = 0 + 10 \cdot 10 = 100 \, \text{m/s} \]

**Hint:** The velocity increases linearly with time under constant acceleration.

### Step 4: Calculate the time taken for the sound to reach the receiver
Since the source is 500 m away from the receiver at `t = 10 seconds`, we need to find out how long it takes for the sound to travel this distance. The speed of sound in air is given as `v = 340 m/s`.

Using the formula:
\[ t_{sound} = \frac{d}{v} = \frac{500}{340} \approx 1.47 \, \text{s} \]

**Hint:** The time taken for sound to travel a distance is calculated using the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \).

### Step 5: Determine the effective time when the sound was emitted
The sound that reaches the receiver at `t = 10 seconds` was emitted at:
\[ t_{emit} = 10 - t_{sound} \approx 10 - 1.47 \approx 8.53 \, \text{s} \]

**Hint:** The sound emitted must account for the time it takes to travel to the receiver.

### Step 6: Calculate the velocity of the source at the time of emission
Now, we need to find the velocity of the source at `t = 8.53 seconds`:
\[ v = u + at = 0 + 10 \cdot 8.53 \approx 85.3 \, \text{m/s} \]

**Hint:** Use the same kinematic equation as before to find the velocity at a specific time.

### Step 7: Apply the Doppler Effect formula
Since the source is moving away from the receiver, we use the Doppler Effect formula:
\[ f' = \frac{v}{v + v_s} f_0 \]
Where:
- \( f' \) = apparent frequency
- \( v \) = speed of sound = 340 m/s
- \( v_s \) = speed of source = 85.3 m/s
- \( f_0 \) = original frequency = 1700 Hz

Substituting the values:
\[ f' = \frac{340}{340 + 85.3} \cdot 1700 \]

### Step 8: Calculate the apparent frequency
Calculating the denominator:
\[ 340 + 85.3 = 425.3 \]
Now substituting into the formula:
\[ f' = \frac{340}{425.3} \cdot 1700 \approx 0.798 \cdot 1700 \approx 1350 \, \text{Hz} \]

**Hint:** When calculating the apparent frequency, ensure to multiply the fraction by the original frequency to find the final result.

### Final Answer
The apparent frequency recorded by the stationary receiver at \( t = 10 \, \text{s} \) will be approximately **1350 Hz**.