Oscillations and Waves Questions and Answers

Answer: C

The group velocity is \(\frac{\Delta\omega}{\Delta k}\).

Adding the two waves gives a carrier at the mean frequency, modulated by an envelope:

\[ \begin{aligned} y &= 2a\cos\left(\frac{\Delta\omega\, t - \Delta k\, z}{2}\right) \\ &\quad \times \cos(\bar{\omega} t - \bar{k} z) \end{aligned} \]

The envelope, which carries the energy, moves with speed \(\frac{\Delta\omega/2}{\Delta k/2} = \frac{\Delta\omega}{\Delta k}\), which is the group velocity \(d\omega/dk\) in the limit of small differences.

The options built from \(\omega \pm \Delta\omega\) over \(k \pm \Delta k\) are phase velocities, of one wave or of the carrier, not the speed of the envelope.

Answer: C

The new time period is \(\sqrt{2}T\).

For a simple pendulum the period depends on the square root of the length:

\[ T = 2\pi\sqrt{\frac{l}{g}} \]

\[ \begin{aligned} T' &= 2\pi\sqrt{\frac{2l}{g}} \\ &= \sqrt{2}\cdot 2\pi\sqrt{\frac{l}{g}} = \sqrt{2}T \end{aligned} \]

Doubling the length multiplies the period by \(\sqrt{2}\), about 1.41. The options \(2\pi T\) and \(\frac{1}{2\pi}T\) wrongly bring the constant \(2\pi\) into the ratio, where it cancels. \(\frac{1}{\sqrt{2}}T\) inverts the relation: a longer pendulum swings more slowly, not faster.

Answer: D

\(\Delta x\) is about 1 cm.

At equilibrium the spring force balances the weight, \(k\,\Delta x = mg\), so \(m/k = \Delta x/g\). The period of the vertical oscillation is then

\[ \begin{aligned} T &= 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{\Delta x}{g}} \\ \Delta x &= \frac{gT^{2}}{4\pi^{2}} = \frac{9.8 \times 0.04}{39.48} \\ &= 0.0099\ \text{m} \approx 1\ \text{cm} \end{aligned} \]

The mass need not be known: the static stretch alone sets the period, exactly as the length sets the period of a pendulum. The options 2, 3 and 4 cm would give periods of about 0.28, 0.35 and 0.40 s.

Answer: D

The correct match is P → 3, Q → 4, R → 5, S → 2.

The detector hears maximum amplitude when the two routes differ in length by a whole number of wavelengths. The smallest \(l\) uses the difference equal to one \(\lambda = 0.29\) m.

(P) The semicircular route has length \(\frac{\pi l}{2}\) against \(l\): \((\frac{\pi}{2} - 1)l = 0.571\,l = 0.29\), so \(l = 0.51\) m (3).

(Q) The rectangular route is \(0.5l + l + 0.5l = 2l\) against \(l\): \(l = 0.29\) m (4).

(R) The route is a side \(l\) plus a semicircle on the diagonal \(\sqrt{2}\,l\), that is \(l + \frac{\pi\sqrt{2}}{2}l = 3.22\,l\). The difference is \(2.22\,l = 0.29\), so \(l = 0.13\) m (5).

(S) In the triangle the angles are 45°, 105° and 30°. By the sine rule the two sides are \(\frac{l\sin 30^{\circ}}{\sin 105^{\circ}}\) and \(\frac{l\sin 45^{\circ}}{\sin 105^{\circ}}\), with \(\sin 105^{\circ} = \cos 15^{\circ} = 0.97\). The route is \(1.244\,l\), the difference is \(0.244\,l = 0.29\), and \(l = 1.19\) m (2).

Answer: E

The time period is 16 s.

Starting from the mean position, \(x = A\sin\omega t\). After 2 s the displacement is \(\frac{A}{\sqrt{2}}\):

\[ \begin{aligned} \sin 2\omega &= \frac{1}{\sqrt{2}} \Rightarrow 2\omega = \frac{\pi}{4} \\ \omega &= \frac{\pi}{8},\quad T = \frac{2\pi}{\omega} = 16\ \text{s} \end{aligned} \]

The particle has moved through one eighth of a cycle (from 0 to \(\frac{\pi}{4}\) of phase), so 2 s is \(\frac{T}{8}\). 8 s would correspond to a quarter cycle, reaching the full amplitude.

Answer: E

The phase difference is \(\pi\).

The bigger (longer) pendulum B has period \(\frac{5T}{2}\), so it completes one oscillation at \(t = \frac{5T}{2}\). In that time A completes

\[ \frac{5T/2}{T} = 2.5 \text{ oscillations} \]

B is back at its starting phase (a whole cycle), while A is half a cycle beyond a whole number of cycles. The phase difference is \(0.5 \times 2\pi = \pi\): A is at the mean position moving the opposite way to B. The smaller fractions of \(\pi\) do not match half a cycle.

Answer: B

The relation is \(\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}\).

For a string under the same tension and with the same mass per unit length, the fundamental frequency is \(n = \frac{1}{2l}\sqrt{\frac{T}{\mu}}\), so \(l = \frac{k}{n}\) with the same constant \(k\) for every segment. Since the lengths add, \(l = l_{1} + l_{2} + l_{3}\):

\[ \frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3} \]

Dividing by \(k\) gives the answer. Adding the frequencies themselves would make the long string's note higher than each piece's, the reverse of the truth; the square-root forms would apply to a quantity varying as \(\frac{1}{\sqrt{l}}\) or \(\sqrt{l}\).

Answer: E

The correct statement is that P.E. is maximum when x is maximum.

In linear SHM, \(PE = \frac{1}{2}kx^{2}\) and \(KE = \frac{1}{2}k(A^{2} - x^{2})\), with total energy \(\frac{1}{2}kA^{2}\) constant. The potential energy is greatest at the extreme positions \(x = \pm A\), where the particle momentarily stops.

At \(x = 0\) the kinetic energy is maximum, not minimum, and the potential energy is zero, not maximum. The total energy is the same everywhere and never zero for an oscillating particle, and the kinetic energy is zero, not maximum, at maximum displacement.

Answer: C

The ratio is 5 : 6.

A closed pipe supports only odd harmonics, \(\frac{v}{4L}, \frac{3v}{4L}, \frac{5v}{4L}, \ldots\), so its third mode is \(\frac{5v}{4L}\). An open pipe supports all harmonics, \(\frac{v}{2L}, \frac{2v}{2L}, \frac{3v}{2L}, \ldots\), so its third mode is \(\frac{3v}{2L}\).

\[ \frac{5v/4L}{3v/2L} = \frac{5}{6} \]

The other ratios do not match these two frequencies. Taking the closed pipe's third harmonic \(3v/4L\) instead of its third mode would give 1 : 2, and 3 : 5 compares the third and fifth harmonics of one pipe.

Answer: B

The speed of sound in steel is \(6000\ \text{m s}^{-1}\).

A rod clamped at its middle has a node at the clamp and antinodes at its two free ends. In the fundamental mode the whole rod is half a wavelength, node in the middle:

\[ \begin{aligned} L &= \frac{\lambda}{2} \Rightarrow \lambda = 2 \times 1 = 2\ \text{m} \\ v &= f\lambda = 3000 \times 2 = 6000\ \text{m s}^{-1} \end{aligned} \]

Taking the rod as a quarter wavelength (as if clamped at one end) gives 12000, not offered. Taking \(\lambda = L\) gives 3000, and the 2000, 4000, 8000 and 9000 values do not follow from either end condition.