Chapter 20: Problem 48

Question:        A tube of length $L=1.18\ {\rm m}$ is closed at one end. A stretched wire is placed near the open end. The wire is $d=0.332\ {\rm m}$ long and has a mass of $m=9.57\times 10^{-3}\ {\rm kg}$. It is fixed at both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration in its fundamental frequency by resonance. Find

1.

the frequency of oscillation of the air column.

2.

the tension in the wire.

Solution:        

For an open ended tube

$$\nu_1 {c\over 4 L}\ =\ 76.27\ {\rm Hz} \ \ \ .$$ = (8)

For the string, we know that $\lambda = 2 d$, and hence $v_{\rm string} = 2 d \nu_1\ =\ 50.9\ {\rm m/s}$. Using the relation between tension, mass per unit length and velocity of waves, we have that

$$\mu v_{\rm string}^2\ =\ {9.57 \times 10^{-3}\over 0.332}\ 50.9^2$$ T =

$$74.7\ {\rm N}$$ = (9)