Problem Overview

The reflected ultrasound frequency heard by the bat is approximately 51 kHz. This result comes from applying the Doppler effect twice: once for the outgoing wave reaching the stationary wall, and once for the reflected wave returning to the moving bat. The calculation uses the standard double Doppler shift formula for a moving source/observer reflecting off a stationary reflector.

A bat emits ultrasound at f₀ = 50 kHz while flying toward a wall at vₛ = 3 m/s, with sound speed v = 330 m/s. The wall is stationary, so the frequency increases due to compression of waves on approach and again on return. No options are provided in the query, but the exact solution is derived step-by-step below.

Step-by-Step Solution

Step 1: Frequency at Wall

The bat (moving source) approaches a stationary wall (observer). The formula is:

f_w = f₀ × (v / (v – vₛ))

Substitute values:

f_w = 50 × (330 / (330 – 3)) = 50 × (330 / 327) ≈ 50.46 kHz

Step 2: Frequency Heard by Bat

The wall reflects at f_w (now stationary source), and the bat moves toward it (moving observer). The formula is:

f_r = f_w × ((v + vₛ) / v)

Substitute:

f_r = 50.46 × ((330 + 3) / 330) = 50.46 × (333 / 330) ≈ 50.92 kHz

Combined Formula

f_r = f₀ × ((v + vₛ) / (v – vₛ)) = 50 × (333 / 327) ≈ 50.92 kHz (rounded to 51 for fill-in-the-blank).

Bats Use Ultrasound for Echolocation

Bats use ultrasound for echolocation, emitting high-frequency waves like 50 kHz that reflect off objects such as walls. When a bat flies directly towards a solid wall at 3 m/s with sound speed at 330 m/s, the bat ultrasound Doppler effect shifts the reflected frequency higher. This 50 kHz reflected frequency calculation is crucial for CSIR NET life sciences and physics exams, demonstrating double Doppler shift.

Doppler Effect in Bat Echolocation

The Doppler effect changes perceived frequency when source and observer move relative to each other. For bats, ultrasound reflection off a stationary wall involves two shifts: outgoing waves compress reaching the wall, then reflected waves compress further as the bat approaches. Key phrase “bat emitting ultrasound at 50 kHz” highlights real-world biology applications in navigation.

Detailed Calculation

Using f₀ = 50 kHz, vₛ = 3 m/s, v = 330 m/s:

• Wall frequency: f_w = 50 × (330 / 327) ≈ 50.46 kHz
• Bat hears: f_r = 50 × (333 / 327) ≈ 50.92 kHz (≈51 kHz)

This matches competitive exam patterns, avoiding approximations since vₛ ≪ v.

Exam Relevance for CSIR NET

CSIR NET aspirants encounter Doppler effect bat flying towards wall problems testing formula application. Common errors include single-shift calculations (yielding ~50.46 kHz) or ignoring directionality. Practice verifies 51 kHz as correct.