Question

a tunin fork of frequency 1000Hz is soudned over resonance tube as shown. speed of sound is v=320m/s and area of cross section of tube is 200cm^2 at a level of 100cm a water outlt is attached it has negligible area of cross section. where will first reso occur? when will it occur? when will second resonance occur?

Answer 1

First resonance occurs when the effective air column length is 8 cm, i.e., when the water level is at 92 cm. Second resonance occurs when the effective air column length is 24 cm, i.e., when the water level is at 76 cm.

Answer 2

To find the first and second resonances in a tube, we use the formula for resonance in a tube closed at one end:

$L_{eff} = \frac{(2n-1)\lambda}{4}$

where $n$ is the resonance number (1, 2, ...), and $\lambda$ is the wavelength of the sound.

1. Calculate the wavelength:

   Given $f = 1000$ Hz and $v = 320$ m/s,

   $\lambda = \frac{v}{f} = \frac{320}{1000} = 0.32$ m

2. First resonance (n=1):

   $L_1 = \frac{\lambda}{4} = \frac{0.32}{4} = 0.08$ m = 8 cm

   Since the water outlet is fixed at the 100 cm level, the first resonance will occur when the water level is lowered so that the air column above the water has a length of 8 cm. That is, the water will be at

   $100$ cm - $8$ cm = $92$ cm

3. Second resonance (n=2):

   $L_2 = \frac{3\lambda}{4} = \frac{3 \times 0.32}{4} = 0.24$ m = 24 cm

   The second resonance occurs when the effective air column is 24 cm long, i.e., when the water level is at

   $100$ cm - $24$ cm = $76$ cm

Therefore:

• First resonance: Occurs when the effective air column length is 8 cm; i.e. when water is lowered to 92 cm (from the 100 cm outlet).
• Second resonance: Occurs when the effective air column length is 24 cm; i.e. when water is lowered to 76 cm.