Question 1, Given: The speed of sound in air is 343 m/s. An open vertical tube h

Question

Question 1,

Given: The speed of sound in air is 343 m/s.

An open vertical tube has water in it. A tuning fork vibrates over its mouth. As the water level is lowered in the tube, a resonance is heard when the water level is 402.5 cm below the top of the tube. And again, after the water level is 437.5 cm below the top of the tube a resonance is heard.

What is the frequency of the tuning fork? Answer in units of Hz.

Question 2,

How many nodes are in the tube after the water has reached the second distance from the top of the tube?

Can you please help me to solve this problem by showin the full explaniation!

Thanks in advance for your assistance,

Explanation / Answer

part 1 )

given v = 343 m/s

d1 = 402.5 cm

d2 = 437.5 cm

The spacing between the resonances is onehalf a wavelength, so

lambda = 2 (d2-d1)

lambda = 70 cm

f = v/lambda = 343 / 70 x 10^-2m = 490 Hz

part 2 )

If there are N1 nodes at the first distance, there are N2 = N1 + 1 nodes at the second distance

d1 = lambda(2N1-1)/4 =  lambda(2(N1+1)-3)/4

d1 = lambda ( N2/2 - labda/4)

N2 = 2d1/ lambda + 3/2

N2 = 2*402.5 / 70 + 3/2

N2 = 13 nodes

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