Two sources are said to be coherent if they
Coherence requires that the source light waves have the same frequency (and wavelength) and maintain a constant phase difference over time.
For a transverse wave on a string, the displacement is described by \(y = A sin(kx – \omega t)\). Then which of the following statement is incorrect?
The expression \(y = A sin(kx - \omega t)\) represents a transverse wave travelling in the positive x direction. The displacement of the particles of the string is along the y-axis, but the wave energy moves along the positive x-axis. Thus, statement (3) is incorrect.
If equation of a wave is given by \(y = 4 \sin \left( 0.4\pi x + 4\pi t + \frac{\pi}{3} \right)\) where x and y are in m and t is in second. Then the magnitude of wave velocity is
For a standard wave equation \(y = A \sin(kx + \omega t + \phi)\), the wave velocity is \(v = \frac{\omega}{k}\). Here, \(k = 0.4\pi\) and \(\omega = 4\pi\), so \(v = \frac{4\pi}{0.4\pi} = 10\text{ m/s}\).
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 expression can be written as \(y = P \sin \left( \frac{2\pi Q}{\lambda}t - \frac{2\pi}{\lambda}x \right)\). The angular frequency is \(\omega = \frac{2\pi Q}{\lambda}\). Thus, the 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 Hz, then original frequency of B was
Since reducing the tension in B decreases its frequency to 630 Hz (unison with A), B's original frequency must have been higher than A's. Therefore, \(f_B = 630 + 3 = 633\text{ Hz}\).
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 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 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, \(f_B = 633\text{ Hz}\).
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 \(|f_1 - f_2| = 8\text{ Hz}\).