Question 314: Waves

Question

The pipe open at both ends and pipe closed at one end have same length and both are vibrating in fundamental mode. Air column vibrating in open pipe has resonance frequency $n_1$ and air column vibrating in closed pipe has resonance frequency $\mathrm{n}_2$, then

Accepted Answer

For an air column vibrating in the fundamental mode:
For a pipe open at both ends, the fundamental wavelength is given by:
$$\lambda_1 = 2L$$
Therefore, the frequency is:
$$n_1 = \frac{v}{\lambda_1} = \frac{v}{2L}$$
For a pipe closed at one end (and open at the other), the fundamental mode has:
$$\lambda_2 = 4L$$
Thus, the fundamental frequency is:
$$n_2 = \frac{v}{\lambda_2} = \frac{v}{4L}$$
Comparing the two frequencies:
$$\frac{n_1}{n_2} = \frac{\frac{v}{2L}}{\frac{v}{4L}} = 2$$
This gives:
$$n_1 = 2n_2$$
Thus, the correct answer is Option A.

Answer

$\mathrm{n}_1=2 \mathrm{n}_2$

Answer

$\mathrm{n}_1=\mathrm{n}_2$

Answer

$2 \mathrm{n}_1=\mathrm{n}_2$

Answer

$3 \mathrm{n}_1=4 \mathrm{n}_2$

Waves

The pipe open at both ends and pipe closed at one end have same length and both are vibrating in fundamental mode. Air column vibrating in open pipe has resonance frequency n_1 and air column vibrating in closed pipe has resonance frequency \mathrm{n}_2, then

Choose the correct answer: