JEE Challenger
More from Waves
PhyMedAdv 2023 • Paper 2

Apparent Frequency Detected from Circular Moving Sound Source

Comprehension Passage

S1S_1 and S2S_2 are two identical sound sources of frequency 656 Hz656\text{ Hz}. The source S1S_1 is located at OO and S2S_2 moves anti-clockwise with a uniform speed 42 m s−14\sqrt{2}\text{ m s}^{-1} on a circular path around OO, as shown in the figure. There are three points PP, QQ and RR on this path such that PP and RR are diametrically opposite while QQ is equidistant from them. A sound detector is placed at point PP. The source S1S_1 can move along direction OPOP.

[Given: The speed of sound in air is 324 m s−1324\text{ m s}^{-1}]

When only S2S_2 is emitting sound and it is at QQ, the frequency of sound measured by the detector in Hz is ________.

Question Diagram 1
Official Numerical Answer648

Topics & Concepts

WavesDoppler EffectSound Waves

Step-by-Step Solution

To find the frequency of sound measured by the detector at point PP when source S2S_2 is at point QQ, we use the Doppler effect formula for a moving source and a stationary observer.

1. Position and Velocity Analysis

Let the origin O(0,0)O(0,0) be the center of the circular path of radius RR.

  • Position vector of detector at PP: r⃗P=−Rj^\vec{r}_P = -R\hat{j}

  • Position vector of source S2S_2 at QQ: r⃗Q=Ri^\vec{r}_Q = R\hat{i}

  • Vector from the source QQ to the detector PP: r⃗QP=r⃗P−r⃗Q=−Ri^−Rj^\vec{r}_{QP} = \vec{r}_P - \vec{r}_Q = -R\hat{i} - R\hat{j}

The unit vector pointing from the source to the detector is: e^QP=r⃗QP∣r⃗QP∣=−Ri^−Rj^R2+R2=−12i^−12j^\hat{e}_{QP} = \frac{\vec{r}_{QP}}{|\vec{r}_{QP}|} = \frac{-R\hat{i} - R\hat{j}}{\sqrt{R^2 + R^2}} = -\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j}

Since S2S_2 moves anti-clockwise on the circle with uniform speed vs=42 m s−1v_s = 4\sqrt{2}\text{ m s}^{-1}, its velocity vector at point Q(R,0)Q(R,0) is along the positive yy-direction: v⃗s=42j^ m s−1\vec{v}_s = 4\sqrt{2}\hat{j}\text{ m s}^{-1}

2. Velocity Component along Line of Sight

The component of velocity of the source directed towards the detector is given by: vs,towards=v⃗s⋅e^QPv_{s, \text{towards}} = \vec{v}_s \cdot \hat{e}_{QP}

Substituting the vectors: vs,towards=(42j^)⋅(−12i^−12j^)=−42×12=−4 m s−1v_{s, \text{towards}} = (4\sqrt{2}\hat{j}) \cdot \left( -\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} \right) = -4\sqrt{2} \times \frac{1}{\sqrt{2}} = -4\text{ m s}^{-1}

The negative sign indicates that the source is moving away (receding) from the detector along the line of sight at a speed of vreceding=4 m s−1v_{\text{receding}} = 4\text{ m s}^{-1}.

3. Calculation of Apparent Frequency

Using the Doppler effect formula for a source moving away from a stationary observer: f′=f(vv+vreceding)f' = f \left( \frac{v}{v + v_{\text{receding}}} \right)

Given data:

  • Original frequency, f=656 Hzf = 656\text{ Hz}
  • Speed of sound in air, v=324 m s−1v = 324\text{ m s}^{-1}
  • Receding speed of source, vreceding=4 m s−1v_{\text{receding}} = 4\text{ m s}^{-1}

Substituting the values: f′=656×(324324+4)f' = 656 \times \left( \frac{324}{324 + 4} \right) f′=656×324328f' = 656 \times \frac{324}{328}

Since 656328=2\frac{656}{328} = 2: f′=2×324=648 Hzf' = 2 \times 324 = 648\text{ Hz}

648

PreviousNext