The lengths of two open organ pipes are l and l + Δl(Δl << l ). If v is the speed of sound, find the frequency of beats between them.

Q: The lengths of two open organ pipes are l and l + Δl(Δl << l ). If v is the speed of sound, find the frequency of beats between them.

Sol: Beat frequency $\large = \nu_1 – \nu_2$

$\large = \frac{v}{2l} – \frac{v}{2(l+\Delta l)} $

$\large = \frac{v}{2l}[1-(1+\frac{\Delta l}{l})^{-1}] $

$\large = \frac{v}{2l}[1-1+\frac{\Delta l}{l}] = \frac{v \Delta l}{2l^2}$

A tuning fork of frequency 340 Hz is vibrated just above a cylindrical tube of length 120 cm…

Q: A tuning fork of frequency 340 Hz is vibrated just above a cylindrical tube of length 120 cm. Water is slowly poured in the tube. If the speed of sound in air is 340 m/s. Find the minimum height of water required for resonance. (v = 340 m/s)

Sol: $\large n = p \frac{v}{4 L}$ with p = 1, 3, 5, ……

So length of air column in the pipe

$\large L = \frac{p v}{4 n} $

L = 25p cm with p = 1, 3, 5, ….

i.e., L = 25 cm, 75 cm, 125 cm

Now as the tube is 120 cm, so length of air column must be lesser than 120 cm, i.e., it can be only 25 cm or 75 cm. Further if h is the height of water filled in the tube, L + h = 120 cm or h = 120 – L

So h will be minimum when Lmax =75 cm

(h)min = 120 – 75 = 45 cm.

A tube of certain diameter and length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz…

Q: A tube of certain diameter and length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. Estimate the diameter of the tube. One end of the tube is now closed. Calculate the frequency of resonance for the tube.

Sol: $\large \nu_0 = \frac{v}{2(L+2e)} $

$\large \nu_0 = \frac{v}{2(L + 2\times 0.6 r)} $

$\large 320 = \frac{320}{2(48 + 1.2 r)} $

r = 10/6 cm

D = 2 r = 2(10/6) = 3.33 cm

Now when one end is closed,

$\large \nu_c = \frac{v}{4(L + 0.6 r)} $

A window whose area is 2 m2 opens on a street where the street noise result in an intensity level at the window of 60 dB…

Q: A window whose area is 2 m2 opens on a street where the street noise result in an intensity level at the window of 60 dB. How much ‘acoustic power’ enters the window via sound waves. Now if an acoustic absorber is fitted at the window, how much energy from street will it collect in five hours ?

Sol: Sound Level $\large \beta = 10 log(\frac{I}{I_0})$

$\large 60 = 10 log(\frac{I}{I_0})$

$\large \frac{I}{I_0} = 10^6 $

$\large I = 10^6 I_0 = 10^6 \times 10^{-12} = 10^{-6} W/m^2$

Intensity $\large I = \frac{E}{A t}$

E = I A t

= 10-6 × 2 × 5 × 3600 = 36 × 10-3 J

Find the speed of sound in a mixture of 1 mol of helium and 2 mol of oxygen at 27°C.

Q: Find the speed of sound in a mixture of 1 mol of helium and 2 mol of oxygen at 27°C.

Sol: $\large \gamma_{mix} = \frac{C_{P_{mix}}}{C_{V_{mix}}} $

$\large = \frac{19R/6}{13R/6} = \frac{19}{13}$

$\large M_{mix} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2} $

$\large M_{mix} = \frac{1 \times 4 + 2 \times 32 }{1 + 2} $

$\large = \frac{68}{3} \times 10^{-3} kg/mol $

$\large v = \sqrt{\frac{\gamma_{mix}R T}{M_{mix}}}$

$\large v = \sqrt{\frac{19}{13}\times \frac{8.314 \times 300}{68 \times 10^{-3}/3}}$

≈ 401 m/s

A sonometer wire has a length of 114 cm between two fixed ends. Where should two bridges be placed to divide the wire…

Q:  A sonometer wire has a length of 114 cm between two fixed ends. Where should two bridges be placed to divide the wire into three segments whose fundamental frequencies are in the ratio 1 : 3 : 4 ?

Sol: In case of a given wire under constant tension,

fundamental frequency of vibration $\large n \propto \frac{1}{l}$

$\large l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4} $

= 12: 4 : 3

l1 = 72 cm; l2 = 24 cm; l3 = 18 cm

First bridge is to be placed at 72 cm from one end.

Second bridge is to be placed at 72 + 24 = 96

A wire having a linear mass density 5.0 × 10-3 kg/m is stretched between two rigid supports with a tension of 450 N…

Q: A wire having a linear mass density 5.0 × 10-3 kg/m is stretched between two rigid supports with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 490 Hz. Find the length of the wire.

Sol: Suppose the wire vibrates at 420 Hz in its nth harmonic and at 490 Hz in its (p +1)th harmonic.

$\large \frac{490}{420} = \frac{p+1}{p}$

or , p = 6

$\large 420 = \frac{6}{2l} \sqrt{\frac{450}{5\times 10^{-3}}}$

l = 2.1 m