Question

A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340 ms−1, what is the distance between the cliff and the observer?​

Answer 1

Answer:

272.0 m/sec.

Explanation:

Frequency of pendulum = 5 vibrations/sec.

∴ Time taken = 8/5 = 1.6 sec.

But d = v x t/2 = 340 x 1.6/2

∴ d = 272.0 m/sec.

Answer 2

$\huge\mathfrak{Answer}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Time taken to make 5 vibrations is - 1 sec

Time taken to make 1 vibration will be - 1/5 sec

Time taken to make 8 vibrations will be - 1/5 × 8 sec

Time taken = 8/5 sec

Velocity of the sound in air = 340 m/s¹

Let distance between the cliff and observer be - d m

Formula ➡

V = 2d / t

Solution ➡

➡ d = ( V × t ) / 2

➡ d = ( 340 × 8 ) / ( 5 × 2 )

➡ d = 340 × 8 / 10

➡ d = 34 × 8

➡ d = 272 m

Ans) The distance between the cliff and observer is 272 m

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬