Question

will mark brainliest. The speed of sound is 340 m/s where a tuning fork produces the second resonance position above a closed air column that is 49.8 cm in length. The frequency of the tuning fork is ___ Hz.

Answer 1

Answer:

Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)

Explanation:

Given:

Speed of sound = 340 m/s

Length of resonance position above a closed air column = 49.8 cm = 0.498 m

Find:

Frequency of the tuning fork

Computation:

Frequency of the tuning fork[second resonance] = 3v / 4l

Frequency of the tuning fork[second resonance] = 3(340) / 4(0.498)

Frequency of the tuning fork[second resonance] = 512.04

Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)

Answer 2

Answer:

The frequency is 512 Hz.

Explanation:

speed, v = 340 m/s

length, L = 49.8 cm = 0.498 m

let the frequency is f.

$f =\frac{3 v}{4 L}\\\\f = \frac{3 \times 340 }{4\times 0.498}\\\\f = 512 Hz$