What is the frequency of the tuning fork?

[rayhan619]

Homework Statement

A 40 cm long tube has 40 cm long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. as the insert is slowly pulled out the sound from the tuning fork creates standing waves in the tube when the total length is 42.5 cm, 56.7 cm, and 70.9 cm. what is the frequency of the tuning fork? The air temperature is 20 degree C.

Homework Equations

f = mv/2L

figure is attached

The Attempt at a Solution

f = 14.2
v = 343 m/s
L =(40+40) cm = 80 cm = 0.8 m

f = mv/2L = (1*343 m/s)/2*0.8 m = 214. 375 hz

got it wrong. not sure how to do it

 

[rl.bhat]

total length is 42.5 cm, 56.7 cm, and 70.9 cm.
These values indicate that the open tube resonates at every increase of equal length of the tube. At open tube this length must be half the wave length.

 

[nlightner]

The frequency of the tuning fork can be calculated using the formula f = mv/2L, where f is the frequency, m is the mode number (1 in this case), v is the speed of sound in air (343 m/s at 20°C), and L is the length of the standing wave in the tube.

In this case, the length of the standing wave in the tube is equal to the length of the tube (40 cm) plus the length of the insert (x cm), which can be pulled out to create different standing wave lengths (42.5 cm, 56.7 cm, and 70.9 cm). So, the equation can be rewritten as:

f = mv/2(40+ x) cm

To solve for x, we can set up a proportion using the given lengths and the corresponding mode numbers:

40 cm / 42.5 cm = 1 / 2
40 cm / 56.7 cm = 2 / 2
40 cm / 70.9 cm = 3 / 2

Solving for x, we get x = 2.5 cm.

Now, substituting this value into the original equation, we get:

f = (1*343 m/s)/2(40+2.5) cm = 214.375 Hz

Therefore, the frequency of the tuning fork is approximately 214.375 Hz.