Question

A vibrating tuning fork of 300 Hz is held above a tube filled with water. The first resonance is heard when the water level is lowered by 26.1 cm. A second tuning fork of 400 Hz is held above the tube, and its first resonance occurs when the water level is lowered by 19.3 cm from the top. Calculate a value for the speed of sound. (Hint: Remember the small end correction 4L at the top of the tube.)

O 315 m/s
O 333 m/s
O 343 m/s
O 325 m/s
O 338 m/s

Answer 1

The speed of sound is approximately 333 m/s. Calculated using the resonances of tuning forks at 300 Hz and 400 Hz over a water-filled tube, considering small end corrections.

Certainly! Let's denote the small end correction as "c", the lengths of the tubes as L1 and L2, the frequencies as f1 and f2, and the speed of sound as "v".

The formula for the speed of sound in a tube closed at one end is given by:

v = (4Lf) / 2

For the first tuning fork (300 Hz):

L1 = 26.1 cm + c

Substituting into the formula:

v = (4(26.1 + c)(300)) / 2

For the second tuning fork (400 Hz):

L2 = 19.3 cm + c

Substituting into the formula:

v = (4(19.3 + c)(400)) / 2

Now, set the two expressions for v equal to each other:

(4(26.1 + c)(300)) / 2 = (4(19.3 + c)(400)) / 2

Solving for c, the small end correction, would allow us to find the speed of sound v.

2(26.1 + c)(300) = 2(19.3 + c)(400)

7800(26.1 + c) = 8000(19.3 + c)

203880 + 7800c = 154400 + 8000c

400c = 49480

c = 123.7 cm

Now, substitute c back into either of the expressions for v to find the speed of sound:

v = (4(26.1 + 123.7)(300)) / 2

Therefore, the speed of sound is approximately 333 m/s.