If two sound waves represented by \(y_1 = 10 \sin(1020\pi t – K_1x)\) and \(y_2 = 10 \sin(1004\pi t – K_2x)\) are superposed at x = 0, then beat frequency is
The linear frequencies are \(f_1 = \frac{1020\pi}{2\pi} = 510\text{ Hz}\) and \(f_2 = \frac{1004\pi}{2\pi} = 502\text{ Hz}\). The beat frequency is \(f_b = |f_1 - f_2| = 8\text{ Hz}\).
If speed of sound in air at \(27^\circ\text{C}\) is \(v\) then at what temperature speed of sound becomes \(2v\)?
The speed of sound \(v \propto \sqrt{T}\. For the speed to double, \(T_2 = 4 T_1 = 4 \times (27 + 273) = 1200\text{ K}\. Converting to Celsius: \(1200 - 273 = 927^\circ\text{C}\).
If two sound waves represented by \(y_1 = 10 \sin(1020\pi t – K_1x)\) and \(y_2 = 10 \sin(1004\pi t – K_2x)\) are superposed at \(x = 0\), then beat frequency is
From \(\omega_1 = 1020\pi\) and \(\omega_2 = 1004\pi\), we get $ f_1 = 510\text{ Hz} $ and \( f_2 = 502\text{ Hz}\). The beat frequency is \(|f_1 - f_2| = 8\text{ Hz}\).
A particle moves according to equation, \(x = a \cos\frac{\pi t}{2}\). The distance covered by it in the time interval between \(t = 0\) to \(t = 3\text{ s}\) is
The time period is \(T = \frac{2\pi}{\pi/2} = 4\text{ s}\). The interval \(t = 3\text{ s}\) corresponds to \(\frac{3T}{4}\). In each quarter cycle, the distance covered is \(a\), so total distance is \(3a\).
If equation of a wave is given by \(y = 4\sin(0.4\pi x + 4\pi t + \frac{\pi}{3})\) where \(x\) and \(y\) are in m and \(t\) is in second.. Then the magnitude of wave velocity is
Wave velocity is given by \(v = \frac{\omega}{k}\). From the wave equation, \(\omega = 4\pi\) and \(k = 0.4\pi\), which gives \(v = \frac{4\pi}{0.4\pi} = 10\text{ m/s}\).
The fifth overtone of a closed pipe is observed to be unison with third overtone of an open pipe. The ratio of the lengths of the pipes is
The frequency of the 5th overtone of a closed pipe is \(f_c = \frac{11v}{4L_c}\). The 3rd overtone of an open pipe is \(f_o = \frac{4v}{2L_o}\). For unison, \(f_c = f_o ⇒\frac{11}{4L_c} = \frac{2}{L_o} \implies \frac{L_c}{L_o} = \frac{11}{8}\).
The displacement of a travelling wave is given by \(y = P \sin \frac{2\pi}{\lambda}(Qt – x)\), where \(t\) is time and \(x\) is distance and \(\lambda\) is wavelength. The linear frequency of the wave is
The wave equation can be rewritten as \(y = P \sin \left(\frac{2\pi Qt}{\lambda} - \frac{2\pi x}{\lambda}\right)\). Here, angular frequency \(omega = \frac{2\pi Q}{\lambda}\). Thus, linear frequency is \(f = \frac{\omega}{2\pi} = \frac{Q}{\lambda}\).
In a guitar, two strings \(A\) and \(B\) are slightly out of tune and produce 3 beats per second. When tension in \(B\) is slightly reduced, both the strings come in unison. If frequency of \(A\) is \(630\text{ Hz}\), then original frequency of \(B\) was
Beats = \(|f_A - f_B| = 3\text{ Hz}\). Since reducing tension in B decreases its frequency to unison (630 Hz), B's initial frequency must have been higher than 630 Hz. Thus, \(f_B = 633\text{ Hz}\).
Which of the following equations represents a progressive harmonic wave travelling along y-axis? (All symbols have their usual meaning)
A progressive wave along the y-axis must have a phase containing the term \((ky \pm \omega t)\). Thus, \(x = A sin(ky - \omega t)\) represents a transverse wave travelling along the y-axis.
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 \(\Delta x = \frac{\lambda}{2\pi} \Delta \phi = \frac{0.66}{2\pi} \left(\frac{\pi}{6}\right) = 0.055\text{ m}\).