Question

A boy sets up a simple pendulum which makes four oscillations per second. The boy then stands in front of a high wall and blows a whistle. He hears an echo, after 6 oscillations of the pendulum. If the speed of sound in air is 330 ms⁻¹, what is the distance between the boy and the wall?

Answer 1

Given:

Frequency of the pendulum = 4 oscillations per second

Number of oscillations before hearing the echo = 6 oscillations

Speed of sound (v) = 330 ms⁻¹

Calculation:

\[ \text{Time for 1 oscillation} = \frac{1}{\text{Frequency}}\]

\[= \frac{1}{4} \ \text{seconds}\]

= 0.25 s

Since the echo is heard after 6 oscillations:

t = 6 × 0.25 s

= 1.5 s

The sound travels to the wall and back, covering a total distance of \(2d\):

2d = v × t

2d = 330 × 1.5

2d = 495

\[ d = \frac{495}{2}\]

d = 247.5 m