The two nearest harmonics of an open organ pipe are 300 Hz and 450 Hz. If speed of sound in the pipe is 300 m/s, then length of the pipe is
For an open organ pipe, successive harmonics differ by the fundamental frequency: \( f_1 = 450 - 300 = 150\text{ Hz} \). Using \( f_1 = \frac{v}{2L} \), we get \( 150 = \frac{300}{2L} \implies L = 1\text{ m} = 100\text{ cm} \).
A progressive wave of frequency \(500\text{ Hz}\) is travelling with a speed of \(330\text{ m/s}\) in air. The distance between the two points which have a phase difference \(30^{circ}\) is
Wavelength \(\lambda = \frac{v}{f} = \frac{330}{500} = 0.66\text{ m}\). Path difference \(\Delta x = \frac{\lambda}{2\pi} \Delta \phi = \frac{0.66}{2\pi} \left(\frac{\pi}{6}\right) = 0.055\text{ m}\).
Two closed pipes produce \(10\text{ beats per second}\) when emitting their fundamental modes. If their lengths are in ratio of \(25 : 26\). Then their fundamental frequencies in Hz are
Frequency \(f \propto \frac{1}{L}\), so \(\frac {f_1}{f_2} = \frac{26}{25}\). Since \(\f_1 -\f_2 = 10\), we have \(26x - 25x = 10 \Rightarrow x = 10\). Thus \(f_1 = 260\text{ Hz}\) and \(f_2 = 250\text{ Hz}\).
In a resonance tube at room temperature two successive resonance lengths of air column are \(25\text{ cm}\) and \(80\text{ cm}\). If the frequency of tuning fork is \(340\text{ Hz}\) then the speed of sound at that temperature is
The speed of sound in a resonance tube is given by \(v = 2 f (l_2 - l_1)\). Substituting \(f = 340\text{ Hz}\), \(l_1 = 0.25\text{ m}\), and \(l_2 = 0.80\text{ m}\), we get \(v = 2(340)(0.80 - 0.25) = 374\text{ m/s}\).