Small Oscillations and Normal Modes Questions and Answers

Answer: B

The statement that does NOT apply is the normal coordinates are orthogonal to one another.

Normal coordinates are the combinations of the original coordinates that diagonalise both the kinetic and potential energies at once. In terms of them,

\[ \begin{aligned} T &= \tfrac{1}{2}\sum_{k}\dot{\eta}_{k}^{2} \\ V &= \tfrac{1}{2}\sum_{k}\omega_{k}^{2}\eta_{k}^{2} \end{aligned} \]

so the potential energy is a sum of squares with constant coefficients, each coordinate oscillates independently, and each has its own eigen-frequency \(\omega_{k}\). Those three statements are the defining properties.

Orthogonality is not a property of the coordinates themselves. The mode (eigen)vectors from which they are built are orthogonal only with respect to the kinetic-energy (mass) matrix, not in the ordinary sense, so the key treats this as the statement that does not apply.

Answer: A

The solution grows as x(t) ∝ t sin(ωt).

The driving frequency equals the natural frequency, so the system is at resonance. Try x = Ct sin ωt:

\[ \begin{aligned} \ddot{x} &= 2C\omega\cos\omega t - C\omega^{2}t\sin\omega t \\ \ddot{x} + \omega^{2}x &= 2C\omega\cos\omega t \Rightarrow C = \frac{F}{2\omega} \end{aligned} \]

This particular solution also satisfies x(0) = 0 and ẋ(0) = 0, so no homogeneous part is needed: x(t) = (F/2ω) t sin ωt, an oscillation whose amplitude grows linearly in time.

t cos ωt does not satisfy ẋ(0) = 0 and does not solve the equation. The amplitude grows without bound but x(t) is finite at every time, so 'x(t) = ∞' is wrong, and there is no exponential growth for an undamped linear oscillator.

Answer: A

The normal-mode frequencies are \(\sqrt{2}\) and \(\sqrt{8}\).

With displacements \(x_{1}, x_{2}\), the equations of motion are

\[ \begin{aligned} m\ddot{x}_{1} &= -k_{1}x_{1} - k_{2}(x_{1} - x_{2}) \\ m\ddot{x}_{2} &= -k_{1}x_{2} - k_{2}(x_{2} - x_{1}) \end{aligned} \]

In the symmetric mode (\(x_{1} = x_{2}\)) the middle spring is not stretched, so \(\omega^{2} = \frac{k_{1}}{m} = 2\). In the antisymmetric mode (\(x_{1} = -x_{2}\)) the middle spring is stretched twice as much, so \(\omega^{2} = \frac{k_{1} + 2k_{2}}{m} = 2 + 6 = 8\).

\(\sqrt{6}\) forgets the outer springs in the second mode (it is \(2k_{2}\) alone), and the options with \(\sqrt{3}\) use \(k_{2}\) for the in-phase mode, where the middle spring plays no part.

Answer: C

The frequency is unchanged: \(\omega = \omega_{0}\).

For small oscillations a simple pendulum has

\[ \omega_{0} = \sqrt{\frac{g}{l}} \]

which depends only on the length and the acceleration due to gravity. The mass cancels, because both the restoring force \(mg\sin\theta\) and the inertia \(ml\) are proportional to m. Halving the mass, with the length the same, therefore leaves the frequency as it was.

Doubling, halving or multiplying by \(\sqrt{2}\) would all require the frequency to depend on m, which it does not; such mass dependence belongs to a spring-mass oscillator, where \(\omega = \sqrt{k/m}\).

Answer: A

The probability is maximum close to ± A.

The time spent in a small interval \( dx \) is \( dx/|v| \), and for SHM the speed is \( v = \omega\sqrt{A^{2} - x^{2}} \). So

\[ P(x)\,dx = \frac{dx}{\pi\sqrt{A^{2} - x^{2}}} \]

which is smallest at \( x = 0 \), where the particle moves fastest, and rises steeply near the turning points \( \pm A \), where it momentarily stops.

So the probability is a minimum, not a maximum, at zero; it is not constant over [−A, A]; and nothing special happens at ±A/2.

Answer: A

For critical damping \(R^{2} = 4L/C\).

The charge in a series LCR circuit obeys \(L\ddot{q} + R\dot{q} + q/C = 0\), with characteristic equation \(Ls^{2} + Rs + 1/C = 0\). Critical damping is the boundary between oscillation and slow decay, where the discriminant is zero:

\[ \begin{aligned} R^{2} - \frac{4L}{C} &= 0 \\ R &= 2\sqrt{\frac{L}{C}} \end{aligned} \]

If \(R^{2}\) is smaller, the circuit oscillates (underdamped), and if larger, it is overdamped. \(R = \sqrt{L/C}\) is half the critical value, and the other forms do not even have the dimensions of resistance squared.