Assertion (A): Wave velocity is equal to group velocity in a non-dispersive medium.
Reason (R): A non-dispersive medium is one in which the wave velocity is frequency dependent.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
In a non-dispersive medium, the phase velocity (wave velocity) \(v_p\) is constant, meaning it does not depend on frequency. In this case, group velocity \(v_g\) equals \(v_p\). So (A) is true. Reason (R) is false because a non-dispersive medium has wave velocity independent of frequency.
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?
1. Wave is moving along +x axis.
2. The shape of the string at t = 0 is a sine wave.
3. Wave is moving along +y axis.
4. Wavelength of the wave is \(\frac{2\pi}{k}\).
View Answer
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.
A stretched string of length \(1\text{ m}\) fixed at both ends having a mass of \(10^{-4}\text{ kg}\) is under a tension of \(16\text{ N}\). The speed of the transverse wave on the string would be
1. \(400\text{ m/s}\)
2. \(200\text{ m/s}\)
3. \(200\sqrt{2}\text{ m/s}\)
4. \(250\text{ m/s}\)
View Answer
Wave speed on a string is \(v = \sqrt{\frac{T}{\mu}}\), where \(\mu = \frac{M}{L} = \frac{10^{-4}\text{ kg}}{1\text{ m}} = 10^{-4}\text{ kg/m}\). Substituting the values, \(v = \sqrt{\frac{16}{10^{-4}}} = \sqrt{16 \times 10^4} = 400\text{ m/s}\).
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
1. \(\frac{Q}{\lambda}\)
2. \(\frac{2\pi Q}{\lambda}\)
3. \(\frac{\lambda}{Q}\)
4. \(\frac{2Q}{\lambda}\)
View Answer
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
1. 630 Hz
2. 627 Hz
3. 640 Hz
4. 633 Hz
View Answer
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, \(f_B = 630 + 3 = 633\text{ Hz}\).