JAMB 2018 Physics · Question 50

A man hears his echo from a nearby hill 2s after he shouted. If the frequency of his voice is 260Hz and the wavelength is 1.29m, how far away is the hill?

Options

  1. A.330.0m
  2. B.335.4m
  3. C.660.0m
  4. D.670.8m

Answer: B. 335.4m

Explanation written with AI and checked against the answer key

1. Analyze the question

We are given:

  • Time taken for the echo to return (tt) = 2 seconds
  • Frequency of the sound (ff) = 260 Hz
  • Wavelength of the sound (λ\lambda) = 1.29 m
  • We need to find the distance (dd) of the hill from the man.
2. Explain the concept
  • Wave Equation: The speed of sound (vv) is calculated using the formula v=f×λv = f \times \lambda.
  • Echo Principle: An echo occurs when sound travels to an obstacle and reflects back. Therefore, the total distance covered by the sound is 2d2d (distance to the hill and back).
  • Speed Formula: Since speed is distance over time (v=total distancetimev = \frac{\text{total distance}}{\text{time}}), the formula for an echo is v=2dtv = \frac{2d}{t}. Rearranging this to find the distance to the hill: d=v×t2d = \frac{v \times t}{2}.
3. Show the solution

Step 1: Calculate the speed of sound (vv)

v=f×λv = f \times \lambda
v=260 Hz×1.29 mv = 260 \, \text{Hz} \times 1.29 \, \text{m}
v=335.4 m/sv = 335.4 \, \text{m/s}

Step 2: Calculate the distance to the hill (dd) Using the echo formula d=v×t2d = \frac{v \times t}{2}:

d=335.4 m/s×2 s2d = \frac{335.4 \, \text{m/s} \times 2 \, \text{s}}{2}

The "2" in the numerator and denominator cancel out:

d=335.4 md = 335.4 \, \text{m}
4. Identify the answer

Comparing our result to the given options: A. 330.0m B. 335.4m C. 660.0m D. 670.8m

The calculated value matches Option B.

5. Tip

Always remember that for echoes, the sound travels the distance twice. A common mistake students make is forgetting to divide by 2, which would lead you to choose 670.8m (Option D). Always double-check if the question asks for the "total distance traveled" or the "distance to the obstacle."

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