JEE PYQ: Motion in a Plane - Question ID a4cef82689c6 (JEE Main 2019)

In this problem, we analyze the apparent position of a moving sound source (the jet aeroplane) due to the finite speed of sound. The key concept is:

• Wavefront Propagation: Sound travels at speed $v$ in still air. When the aeroplane moves, the wavefronts emitted at different times reach the observer at different angles, creating an illusion of the plane’s position.
• Relative Motion & Apparent Angle: The observer hears the sound coming from an angle $\theta = 60^\circ$ with the ground, but sees the plane directly overhead. This discrepancy arises because the plane has moved during the time sound travels from its earlier position to the observer.
• Key Formula: If the plane moves horizontally at speed $u$, and sound travels at speed $v$, the apparent angle $\theta$ of the sound source satisfies: $\tan \theta = \frac{u}{v}$ This relation comes from the geometry of wavefronts and the relative motion of the source and observer.

Step 1: Define Variables & Geometry

• Let the observer be at point $O$ on the ground.
• At time $t = 0$, the plane is at point $P$, at a horizontal distance $x$ from $O$, and at height $h$.
• The plane moves horizontally with speed $u$ (toward the observer, from north to south).
• Sound emitted at $t = 0$ from $P$ reaches $O$ at time $t = \frac{\sqrt{x^2 + h^2}}{v}$, since sound travels the distance $\sqrt{x^2 + h^2}$ at speed $v$.

Step 2: Position of Plane When Sound Reaches Observer

• During the time $t$, the plane moves a distance $u \cdot t$ toward $O$.
• At time $t$, the plane is directly above $O$, so its horizontal position is $0$. Thus: $x - u \cdot t = 0 \quad \Rightarrow \quad x = u \cdot t$

Step 3: Relate Angle of Sound Arrival to Geometry

• The observer hears the sound coming from an angle $60^\circ$ with the ground. This means the wavefront arrives at an angle $\theta = 60^\circ$, so: $\tan 60^\circ = \frac{h}{x}$ Since $\tan 60^\circ = \sqrt{3}$, we have: $h = x \sqrt{3}$

Step 4: Substitute and Solve for $u$

• From Step 2, $x = u \cdot t$, and from the sound travel time: $t = \frac{\sqrt{x^2 + h^2}}{v}$ Substitute $h = x \sqrt{3}$: $t = \frac{\sqrt{x^2 + 3x^2}}{v} = \frac{2x}{v}$ Now, substitute $t = \frac{2x}{v}$ into $x = u \cdot t$: $x = u \cdot \frac{2x}{v} \quad \Rightarrow \quad 1 = \frac{2u}{v}$ Solving for $u$: $u = \frac{v}{2}$

Step 5: Match with Given Options

• The speed of the plane is $\frac{v}{2}$, which corresponds to option D.

Students often confuse the apparent angle of sound arrival with the actual position of the plane. A frequent mistake is assuming the plane is at the angle from which the sound is heard, leading to incorrect trigonometric relations. Always remember:

• The sound heard at $60^\circ$ is from the plane’s earlier position, not its current position.
• The plane’s speed is derived from the time delay of sound travel and the displacement during that time.
• Drawing a clear diagram with the plane’s initial and final positions is crucial to avoid misinterpreting the geometry.