Question

An open organ pipe of length L vibrates in second harmonic mode. The pressure vibration is maximum

• A. at the two ends
• B. at a distance L/4 from either end inside the tube
• C. at the mid-point of the tube
• D. none of these

Answer 1

Answer: (B)

at a distance L/4 from either end inside the tube

For an open organ pipe, the length L is given as

$L=\frac{n\lambda }{2}$

where, $\lambda$ is the wavelength of wave and n is an integer and by putting n = 1,2,3,............... we get the modes of vibration.

n=1 gives first harmonics, n=2 gives second harmonics and so on.

Here, an open organ pipe of length L vibrates in second harmonic mode,

hence the length of pipe is

$L=\frac{2\lambda }{2}$

$L=\lambda$

Hence, the length of pipe is equal to the distance between two consecutive antinodes at both open ends. The nodes are at distance $\frac{\lambda }{4}$ from the antinodes. Since there is no displacement of particles at nodes the pressure vibration is maximum there.

Hence, in open organ pipe the pressure vibration is maximum at a distance

$\frac{L}{4}$(since, $L=\lambda )$ from either end inside the tube as shown in figure.