Oscillations and Waves Questions and Answers

Answer: E

\(A = \) 4 m.

A sine and a cosine of the same phase are 90° apart, so they combine like perpendicular vectors:

\[ \begin{aligned} R &= \sqrt{3^{2} + A^{2}} = 5 \\ A^{2} &= 25 - 9 = 16,\quad A = 4\ \text{m} \end{aligned} \]

2 m comes from simply subtracting amplitudes (5 - 3), which would be right only for waves in phase.

Formally, \(3\sin\theta + A\cos\theta = R\sin(\theta + \delta)\) with \(R = \sqrt{9 + A^{2}}\), which is the resultant amplitude.

Answer: A

The fundamental frequency of the open pipe is 100 Hz.

For a pipe of length \(L\): open at both ends, \(f_{0} = \frac{v}{2L}\); closed at one end, the harmonics are odd multiples of \(\frac{v}{4L}\), so the third harmonic is \(\frac{3v}{4L} = \frac{3}{2}f_{0}\).

\[ \begin{aligned} \tfrac{3}{2}f_{0} - f_{0} &= 50 \\ \tfrac{1}{2}f_{0} &= 50 \Rightarrow f_{0} = 100\ \text{Hz} \end{aligned} \]

Check: the closed pipe's third harmonic is 150 Hz, 50 Hz above 100 Hz. Taking the closed pipe's third harmonic as \(3f_{0}\) would give 25 Hz, not offered.

Answer: E

The wave velocity is 200 m s^-1.

A travelling wave of the form \(y = A\sin\frac{2\pi}{\lambda}(x - vt)\) moves in the positive x-direction with speed \(v\). Comparing with \(y = 0.05\sin\frac{2\pi}{\lambda}(x - 200t)\) gives \(v = 200\) m s^-1 directly.

Equivalently, \(k = \frac{2\pi}{\lambda}\) and \(\omega = \frac{2\pi}{\lambda} \times 200\), so \(v = \frac{\omega}{k} = 200\). The square-root options have no basis in this equation, and 400 doubles the speed.

Answer: A

The unknown fork can be 260 Hz or 252 Hz.

The beat frequency equals the difference between the two frequencies. Four beats per second means the unknown frequency differs from 256 Hz by 4 Hz, so it is 256 + 4 = 260 Hz or 256 - 4 = 252 Hz. Which one it is cannot be told without a further test, such as loading one fork with wax.

Each other pair includes a value that does not differ from 256 Hz by 4: 256 Hz itself would give no beats, and 258, 250, 248, 255, 257 and 215 Hz would give 2, 6, 8, 1, 1 and 41 beats per second.

Answer: B

Simple harmonic motion is represented by a = -10x.

SHM is defined by an acceleration proportional to the displacement and directed opposite to it: \(a = -\omega^{2}x\). Here \(\omega^{2} = 10\), so \(\omega = \sqrt{10}\) rad/s.

a = 1.2x is proportional but points away from equilibrium, so the motion runs away instead of oscillating. \(a = 0.9x^{2}\) and \(a = 100x^{3}\) are not proportional to x, and \(a^{2} = 12x\) is not even linear in a.

Answer: C

The wave speed is 80 m s^-1.

The speed of a transverse wave on a string is \(v = \sqrt{T/\mu}\), where \(\mu\) is the mass per unit length.

\[ \begin{aligned} \mu &= \frac{0.005\ \text{kg}}{0.5\ \text{m}} = 0.01\ \text{kg/m} \\ v &= \sqrt{\frac{64}{0.01}} = \sqrt{6400} = 80 \end{aligned} \]

Dividing 5 g by 50 cm without converting units gives \(\mu = 0.1\) and \(v = \sqrt{640} = 8\sqrt{10}\), option B. Only the SI values, 0.005 kg over 0.5 m, give the right \(\mu\).

Answer: E

The force constant of each piece is \(nk\).

For a spring of given material and coil, the spring constant is inversely proportional to its length: \(k \propto \frac{1}{L}\). Each piece is \(\frac{L}{n}\) long, so its constant is \(n\) times larger.

A check: the n pieces joined end to end in series must give back the original spring, and \(n\) springs of constant \(nk\) in series give \(\frac{nk}{n} = k\).

\(\frac{k}{n}\) has the proportion the wrong way round; a shorter spring is stiffer, not softer. \(nk^{2}\), \(\frac{n^{2}}{k}\) and \(\frac{n}{k}\) do not even have the units of a spring constant.

Answer: E

The angular frequency is 4 s^-1.

In simple harmonic motion the acceleration is \(a = -\omega^{2}x\), so in magnitude:

\[ \begin{aligned} \omega^{2} &= \frac{a}{x} = \frac{48}{3} = 16 \\ \omega &= 4 \text{ s}^{-1} \end{aligned} \]

Forgetting the square root gives 16, which is not an option; 6, 3, 5 and 2 do not satisfy \(\omega^{2} = 16\).

Answer: C

The frequencies are 100 Hz, 300 Hz, 500 Hz, 700 Hz, ...

For an open pipe of length L, the fundamental is \(\frac{v}{2L} = 200\) Hz. Closing one end makes the fundamental \(\frac{v}{4L}\), half of that, so 100 Hz. A pipe closed at one end produces only the odd harmonics, 1, 3, 5, 7 ... times the fundamental: 100, 300, 500, 700 Hz.

The series starting at 200 Hz keep the open-pipe fundamental, and 100, 200, 300, 400 Hz includes even harmonics, which a closed pipe cannot produce.

Answer: A

The amplitude is 4 cm.

A combination \(a\sin\theta + b\cos\theta\) is a single sinusoid of amplitude \(\sqrt{a^{2} + b^{2}}\). Here \(a = 2\) and \(b = 2\sqrt{3}\):

\[ \begin{aligned} A &= \sqrt{2^{2} + (2\sqrt{3})^{2}} = \sqrt{4 + 12} = 4 \\ y &= 4\sin(2\pi t + \tfrac{\pi}{3}) \end{aligned} \]

2 is just the factor outside the bracket, and adding the coefficients directly does not give an amplitude.