MCQ
An open organ pipe of length $L$ vibrates in second harmonic mode. The pressure vibration is maximum
- Aat the two ends
- ✓at a distance $L/4$ from either end inside the tube
- Cat the mid-point of the tube
- Dnone of these
✓
Answer
Correct option: B.
at a distance $L/4$ from either end inside the tube
b
For an open organ pipe, the length $L$ is given as
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
$\frac{L}{4}(\text { since }, L=\lambda)$ from either end inside the tube as shown in figure.
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