At t=0 , observer and source are at same place. Now the source is projected with velocity `60 sqrt2 `m/s at `45 ^(@)` . Natural frequency of source is 1000 Hz.find the frequency heard by the observer at t=2s. Take speed of sound = 340 m/s

Options

A. 930 Hz
B. 860 Hz
C. 826 Hz
D. 970Hz

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the components of the source's velocity
The source is projected with a velocity of \(60\sqrt{2}\) m/s at an angle of \(45^\circ\). We can find the horizontal and vertical components of this velocity.

- Horizontal component (\(V_x\)):
\[ V_x = 60\sqrt{2} \cdot \cos(45^\circ) = 60\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 60 \text{ m/s} \]

- Vertical component (\(V_y\)):
\[ V_y = 60\sqrt{2} \cdot \sin(45^\circ) = 60\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 60 \text{ m/s} \]

### Step 2: Calculate the position of the source after 2 seconds
Using the equations of motion, we can find the position of the source after \(t = 2\) seconds.

- Horizontal distance (\(x\)):
\[ x = V_x \cdot t = 60 \cdot 2 = 120 \text{ m} \]

- Vertical distance (\(y\)) considering the effect of gravity:
\[ y = V_y \cdot t - \frac{1}{2} g t^2 = 60 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2^2) = 120 - 20 = 100 \text{ m} \]

Thus, the position of the source after 2 seconds is:
\[ \text{Position of source} = (120 \hat{i} + 100 \hat{j}) \text{ m} \]

### Step 3: Calculate the distance from the observer to the source
The observer is at the origin \((0, 0)\). The distance \(r\) from the observer to the source is given by:
\[ r = \sqrt{x^2 + y^2} = \sqrt{120^2 + 100^2} = \sqrt{14400 + 10000} = \sqrt{24400} = 20\sqrt{61} \text{ m} \]

### Step 4: Calculate the effective velocity of the source
The effective velocity of the source in the direction of the observer can be calculated using the dot product of the source's velocity vector and the unit vector in the direction of the observer.

- The unit vector in the direction of the observer:
\[ \hat{r} = \frac{(120 \hat{i} + 100 \hat{j})}{20\sqrt{61}} = \left(\frac{120}{20\sqrt{61}}, \frac{100}{20\sqrt{61}}\right) = \left(\frac{6}{\sqrt{61}}, \frac{5}{\sqrt{61}}\right) \]

- The effective velocity \(V_{eff}\):
\[ V_{eff} = V_x \cdot \frac{6}{\sqrt{61}} + V_y \cdot \frac{5}{\sqrt{61}} = 60 \cdot \frac{6}{\sqrt{61}} + 60 \cdot \frac{5}{\sqrt{61}} = \frac{60(6 + 5)}{\sqrt{61}} = \frac{660}{\sqrt{61}} \text{ m/s} \]

### Step 5: Apply the Doppler Effect formula
The apparent frequency \(f'\) heard by the observer can be calculated using the Doppler Effect formula:
\[ f' = f_0 \cdot \frac{v + v_o}{v - v_s} \]
Where:
- \(f_0 = 1000 \text{ Hz}\) (natural frequency of the source)
- \(v = 340 \text{ m/s}\) (speed of sound)
- \(v_o = 0 \text{ m/s}\) (observer is at rest)
- \(v_s = V_{eff}\)

Substituting the values:
\[ f' = 1000 \cdot \frac{340 + 0}{340 - \frac{660}{\sqrt{61}}} \]

Calculating the denominator:
\[ 340 - \frac{660}{\sqrt{61}} \approx 340 - 84.5 \approx 255.5 \]

Now substituting back:
\[ f' = 1000 \cdot \frac{340}{255.5} \approx 1000 \cdot 1.33 \approx 826 \text{ Hz} \]

### Final Answer
The frequency heard by the observer at \(t = 2\) seconds is approximately **826 Hz**.

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