Question

A pipe closed at one end has length 42.5 cm. Calculate the fundamental frequency if the speed of sound in air is 340 m/s.​

Answer 1

Answer:

429.165hz

Step-by-step explanation:

l=42.5

v= 340m/s

d= 3.4

=3.4×10^2m/s

0.034m

n=v/4l

n=340/4×0.459

n=39.015

np=(2p+1)n

n5=(2×5h)n

n5=11n

=11×39.015

=429.165hz

Answer 2

Answer:

The pipe has a fundamental frequency of 800 Hz

Step-by-step explanation:

Given that the Length of the tube = 42.5cm

and that the Speed of sound in air = 340m/s

We need to find the fundamental frequency

We know that for a closed tube, fundamental frequency = $\frac{V}{4L}$

where V is the speed of sound in air

and L is the length of the pipe.

Therefore, on substituting the values, we get

Fundamental Frequency = $\frac{340}{0.425}$   (as 1 m = 100 cm)

                                          = 800

Therefore, the pipe has a fundamental frequency of 800 Hz