The Workbook

1. \(\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{100}{1}} = 10\) rad/s.

2. \(T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{10} = \dfrac{\pi}{5} \approx 0.63\) s.

3. \(f = \dfrac{1}{T} = \dfrac{5}{\pi} \approx 1.6\) Hz.

1. At rest \(kx_0 = mg\): \(k = \dfrac{1 \times 10}{0.10} = 100\) N/m.

2. \(T = 2\pi\sqrt{\dfrac{m}{k}}\) and \(\dfrac{m}{k} = \dfrac{x_0}{g}\) from \(kx_0 = mg\). So \(T = 2\pi\sqrt{\dfrac{x_0}{g}} = 2\pi\sqrt{\dfrac{0.1}{10}} = \dfrac{2\pi}{10} = \dfrac{\pi}{5}\) s.

   The static stretch alone fixes the period; the mass and stiffness need not be known separately.

\(T = 2\pi\sqrt{m/k}\).

1. \(m \to 2m\): \(T \to \sqrt{2}\,T\).

2. \(k \to 2k\): \(T \to T/\sqrt{2}\).

3. The period does not depend on the amplitude: still \(T\).

1. Series: both springs carry the same force \(F\) and each stretches by \(F/k\); the total stretch is \(2F/k\), so \(k_s = \dfrac{k}{2} = 50\) N/m.

   Parallel: both springs stretch by the same \(x\) and each pulls with \(kx\); the total force is \(2kx\), so \(k_p = 2k = 200\) N/m.

2. \(T \propto \dfrac{1}{\sqrt{k}}\): \(\dfrac{T_s}{T_p} = \sqrt{\dfrac{k_p}{k_s}} = \sqrt{\dfrac{200}{50}} = 2\).

For a given force each coil stretches by the same amount, so the total stretch is proportional to the length. The stiffness is therefore inversely proportional to the length: \(k \propto \dfrac{1}{L}\).

The shorter part (\(\dfrac{1}{3}\) of the length) has \(3k = 300\) N/m; the longer part (\(\dfrac{2}{3}\)) has \(\dfrac{3k}{2} = 150\) N/m. Check: in series they give \(\dfrac{300 \times 150}{450} = 100\) N/m.

1. Moving right by \(x\) stretches \(k_1\) and compresses \(k_2\). Both push the block back: \(F = -k_1x - k_2x = -(k_1 + k_2)x = -400x\).

   The springs act like a parallel pair: \(k_{\text{eff}} = 400\) N/m.

2. \(T = 2\pi\sqrt{\dfrac{1}{400}} = \dfrac{2\pi}{20} = \dfrac{\pi}{10} \approx 0.31\) s.

1. Along the incline: \(kx_0 = mg\sin 30^\circ = 5\) N, so \(x_0 = \dfrac{5}{100} = 0.05\) m \(= 5\) cm.

2. Gravity only shifts the equilibrium; the restoring force for a displacement from it is still \(-kx\). \(T = 2\pi\sqrt{\dfrac{1}{100}} = \dfrac{\pi}{5}\) s, the same as on a horizontal floor.

1. The stretch varies between \(x_0 - A = 5\) cm and \(x_0 + A = 15\) cm.

   \(T_{\max} = k(x_0 + A) = 100 \times 0.15 = 15\) N at the lowest point; \(T_{\min} = 100 \times 0.05 = 5\) N at the highest point.

2. The spring is slack when the stretch is zero at the top: \(A = x_0 = 10\) cm. For a larger amplitude the block would leave the spring's control near the top.

1. No external horizontal force acts, so the centre of mass stays at rest. It is \(3\) times closer to the \(3\) kg block, so the \(1\) kg block moves \(3\) times as far.

2. \(\mu = \dfrac{1 \times 3}{1 + 3} = 0.75\) kg. \(T = 2\pi\sqrt{\dfrac{0.75}{300}} = 2\pi\sqrt{\dfrac{1}{400}} = \dfrac{\pi}{10} \approx 0.31\) s.

No. The period depends only on \(m\) and \(k\); the acceleration of the lift changes neither.

In the lift the effective gravity is \(12\) m/s\(^2\), so the equilibrium stretch is \(\dfrac{1 \times 12}{100} = 12\) cm, that is, \(2\) cm lower than before. The block oscillates about this new point.

1. Together the blocks have mass \(4\) kg: \(\omega = \sqrt{\dfrac{400}{4}} = 10\) rad/s.

2. Only friction acts horizontally on the upper block, so friction supplies its restoring force \(ma\). The largest friction available is \(\mu mg = 0.5 \times 1 \times 10 = 5\) N, giving \(a_{\max} = 5\) m/s\(^2\).

   The upper block follows without slipping while \(\omega^2 A \le 5\): \(A \le \dfrac{5}{100} = 0.05\) m \(= 5\) cm.

1. \(\omega^2 = \dfrac{100}{1} = 100\). The largest downward acceleration of the platform is \(\omega^2 A = 100 \times 0.05 = 5\) m/s\(^2\), less than \(g\). The platform never falls away faster than the block would fall, so the block stays on.

2. The block leaves where the platform's downward acceleration reaches \(g\): \(\omega^2 x = 10\), \(x = 0.1\) m \(= 10\) cm above the equilibrium position, on the way up. With \(A = 15\) cm this point is reached.

\(E\) is constant, so \(\dfrac{dE}{dt} = 0\): \(mv\dfrac{dv}{dt} + kx\dfrac{dx}{dt} = 0\).

With \(\dfrac{dx}{dt} = v\) and \(\dfrac{dv}{dt} = a\): \(v(ma + kx) = 0\). Since \(v\) is not always zero, \(a = -\dfrac{k}{m}x\).

This is SHM with \(\omega^2 = \dfrac{k}{m}\), so \(T = 2\pi\sqrt{\dfrac{m}{k}}\).

1. About the equilibrium position, where \(kx_0 = mg\): \(x_0 = 10\) cm below the release point.

2. The block is released from rest \(10\) cm from the equilibrium position, so that is the amplitude: \(A = 10\) cm.

3. The lowest point is \(A\) below equilibrium: extension \(x_0 + A = 20\) cm, twice the static stretch.

Series: \(k = \dfrac{k_1k_2}{k_1 + k_2} = \dfrac{100 \times 300}{400} = 75\) N/m.

\(T = 2\pi\sqrt{\dfrac{0.75}{75}} = 2\pi\sqrt{0.01} = \dfrac{2\pi}{10} = \dfrac{\pi}{5}\) s.

\(T^2 \propto m\): \(\dfrac{4}{9} = \dfrac{m}{m + 2}\), so \(9m = 4m + 8\) and \(m = 1.6\) kg.

1. \(\omega = \sqrt{\dfrac{200}{2}} = 10\) rad/s. \(v = \omega\sqrt{A^2 - x^2} = 10\sqrt{100 - 36}\) cm/s \(= 80\) cm/s \(= 0.8\) m/s.

2. \(F = kA = 200 \times 0.1 = 20\) N, directed towards the mean position.

Half the spring has twice the stiffness. \(T \propto \dfrac{1}{\sqrt{k}}\), so the new period is \(\dfrac{T}{\sqrt{2}}\).

\([k] = \dfrac{[\text{force}]}{[\text{length}]} = \dfrac{\mathrm{MLT^{-2}}}{\mathrm{L}} = \mathrm{MT^{-2}}\).

\(\left[\sqrt{\dfrac{m}{k}}\right] = \sqrt{\dfrac{\mathrm{M}}{\mathrm{MT^{-2}}}} = \sqrt{\mathrm{T^2}} = \mathrm{T}\), a time. The \(2\pi\) has no dimension.

1. \(v_{\max} = A\omega\): \(A = \dfrac{1}{10} = 0.1\) m \(= 10\) cm.

2. \(x = 5\) cm is \(\dfrac{A}{2}\), reached after \(\dfrac{T}{12}\) from the mean position (phase \(30^\circ\)): \(t = \dfrac{\pi/5}{12} = \dfrac{\pi}{60} \approx 0.05\) s.

1. Four springs in parallel: \(k = 4 \times 2.5 \times 10^{4} = 10^{5}\) N/m. \(T = 2\pi\sqrt{\dfrac{1000}{10^{5}}} = 2\pi \times 0.1 = \dfrac{\pi}{5} \approx 0.63\) s.

2. Extra weight \(= 300 \times 10 = 3000\) N. Extra compression \(= \dfrac{3000}{10^{5}} = 0.03\) m \(= 3\) cm.

\(T = 2\pi\sqrt{\dfrac{I}{mgd}}\), where \(I\) is about the pivot and \(d\) is the distance from the pivot to the centre of mass. For a point bob, \(I = mL^2\) and \(d = L\): \(T = 2\pi\sqrt{\dfrac{mL^2}{mgL}} = 2\pi\sqrt{\dfrac{L}{g}}\).

\(I = \dfrac{mL^2}{3}\), \(d = \dfrac{L}{2}\): \(T = 2\pi\sqrt{\dfrac{mL^2/3}{mgL/2}} = 2\pi\sqrt{\dfrac{2L}{3g}} = 2\pi\sqrt{\dfrac{3}{30}} = \dfrac{2\pi}{\sqrt{10}} \approx 2.0\) s.

1. \(\dfrac{I}{m} = \dfrac{L^2}{12} + d^2 = 0.12 + 0.16 = 0.28\) m\(^2\).

2. \(L_{\text{eq}} = \dfrac{I}{md} = \dfrac{0.28}{0.4} = 0.7\) m.

3. \(T = 2\pi\sqrt{\dfrac{L_{\text{eq}}}{g}} = 2\pi\sqrt{0.07} \approx 1.7\) s.

About the rim, \(I = \dfrac{1}{2}MR^2 + MR^2 = \dfrac{3}{2}MR^2\) and \(d = R\). \(T = 2\pi\sqrt{\dfrac{3R}{2g}} = 2\pi\sqrt{\dfrac{1.8}{20}} = 2\pi\sqrt{0.09} = 0.6\pi \approx 1.9\) s.

\(I = MR^2 + MR^2 = 2MR^2\), \(d = R\). \(T = 2\pi\sqrt{\dfrac{2R}{g}} = 2\pi\sqrt{0.04} = 0.4\pi \approx 1.3\) s.

Displaced by \(\theta\), the weight has lever arm \(d\sin\theta\) about the pivot, giving a restoring torque \(-mgd\sin\theta \approx -mgd\,\theta\). So \(I\dfrac{d^2\theta}{dt^2} = -mgd\,\theta\): simple harmonic motion with \(\omega^2 = \dfrac{mgd}{I}\) and \(T = 2\pi\sqrt{\dfrac{I}{mgd}}\).

The equivalent length is the length of the simple pendulum with the same period: \(2\pi\sqrt{L_{\text{eq}}/g} = 2\pi\sqrt{I/mgd}\) gives \(L_{\text{eq}} = \dfrac{I}{md}\). With \(I = m(k^2 + d^2)\) by the parallel axis theorem: \(L_{\text{eq}} = \dfrac{k^2 + d^2}{d}\). Rod at its end: \(k^2 = \dfrac{L^2}{12}\), \(d = \dfrac{L}{2}\): \(L_{\text{eq}} = \dfrac{L^2/12 + L^2/4}{L/2} = \dfrac{2L}{3}\).

\(L_{\text{eq}} = \dfrac{k^2}{d} + d\). \(\dfrac{dL_{\text{eq}}}{dd} = -\dfrac{k^2}{d^2} + 1 = 0\) at \(d = k\), and the second derivative \(\dfrac{2k^2}{d^3}\) is positive, so it is a minimum. For the rod: \(k = \dfrac{L}{\sqrt{12}} = \dfrac{1.2}{3.46} \approx 0.35\) m.

1. As \(d \to 0\) the pivot approaches the centre of mass and the restoring torque \(mgd\,\theta\) vanishes. With no torque to bring it back, the body takes longer and longer to swing.

2. For large \(d\), \(I \approx md^2\) grows as \(d^2\) while the torque grows only as \(d\). So \(T \approx 2\pi\sqrt{d/g}\) grows like a long simple pendulum.

Equal periods mean equal \(L_{\text{eq}}\): \(\dfrac{k^2}{d_1} + d_1 = \dfrac{k^2}{d_2} + d_2\), so \(k^2\dfrac{d_2 - d_1}{d_1d_2} = d_2 - d_1\) and \(d_1d_2 = k^2\). For the rod, \(k^2 = 0.12\) and \(d_1 = 0.4\): \(d_2 = 0.3\) m. \(d_1 + d_2 = 0.7\) m \(= L_{\text{eq}}\), as in Q3.

For pivots on opposite sides with the same period, \(d_1 + d_2 = L_{\text{eq}}\) (Q10). So \(T = 2\pi\sqrt{\dfrac{d_1 + d_2}{g}}\) and \(g = \dfrac{4\pi^2(d_1 + d_2)}{T^2} = \dfrac{4\pi^2 \times 1}{4} = \pi^2 \approx 9.9\) m/s\(^2\). This is the idea of Kater's pendulum.

\(k^2 = 0.12\) m\(^2\). End: \(d = 0.6\), \(L_{\text{eq}} = \dfrac{0.12 + 0.36}{0.6} = 0.8\) m. Pivot at \(0.3\) m: \(L_{\text{eq}} = \dfrac{0.12 + 0.09}{0.3} = 0.7\) m. The shorter \(L_{\text{eq}}\) gives the shorter period, so the pivot at \(0.3\) m, which is closer to \(k \approx 0.35\) m.

\(T_{\text{bob}} = 2\pi\sqrt{L/g}\) and \(T_{\text{rod}} = 2\pi\sqrt{2L/3g}\). Ratio \(\dfrac{T_{\text{rod}}}{T_{\text{bob}}} = \sqrt{\dfrac{2}{3}} \approx 0.82\). The rod swings faster: its mass is spread nearer the pivot, so its equivalent length is only \(\dfrac{2L}{3}\).

The centre of percussion is the point where a blow produces no impulsive reaction at the pivot. With an impulse \(J\) at distance \(q\) and no pivot impulse: linear, \(J = m\,v_{\text{cm}} = m\omega d\); angular about the pivot, \(Jq = I\omega\). Dividing: \(q = \dfrac{I}{md}\), which is exactly \(L_{\text{eq}}\).

1. \(q = \dfrac{I}{md} = \dfrac{mL^2/3}{mL/2} = \dfrac{2L}{3} = 0.8\) m.

2. \(J_p = J\left(\dfrac{3x}{2L} - 1\right) = 6\left(\dfrac{3.6}{2.4} - 1\right) = 6 \times 0.5 = 3\) N s, in the same sense as \(J\).

3. \(J_p = 6\left(\dfrac{1.8}{2.4} - 1\right) = 6 \times (-0.25) = -1.5\) N s: \(1.5\) N s opposite to \(J\).

Let the pivot supply an impulse \(J_p\) in the direction of \(J\). Linear: \(J + J_p = m\omega\dfrac{L}{2}\). Angular about the pivot: \(Jx = \dfrac{mL^2}{3}\omega\), so \(m\omega = \dfrac{3Jx}{L^2}\). Then \(J_p = \dfrac{3Jx}{L^2}\cdot\dfrac{L}{2} - J = J\left(\dfrac{3x}{2L} - 1\right)\). It vanishes at \(x = \dfrac{2L}{3}\).

1. The sweet spot is the centre of percussion: \(\dfrac{2L}{3} = \dfrac{2 \times 0.9}{3} = 0.6\) m from the hands.

2. At the tip \(x = L\), so \(J_p = J\left(\dfrac{3}{2} - 1\right) = \dfrac{J}{2}\): the hands must supply half the ball's impulse, in the ball's direction. That is the sting felt on a mistimed shot.

The door is a rod pivoted at its hinged edge. The stop should meet it at the centre of percussion, \(\dfrac{2}{3} \times 0.9 = 0.6\) m from the hinge line.

Both distances equal \(\dfrac{I}{md}\). A blow at that point makes the body start moving exactly as a free particle placed there would, so the pivot has nothing to do. In the same way, a bob at that distance swings with the same period as the body, so the pivot's constraint again adds nothing.

Centre of percussion at \(0.8\) m. \(I = \dfrac{2 \times 1.44}{3} = 0.96\) kg m\(^2\). \(Jq = I\omega\): \(3 \times 0.8 = 0.96\,\omega\), so \(\omega = 2.5\) rad/s. Check: \(m\omega d = 2 \times 2.5 \times 0.6 = 3\) N s \(= J\), so the pivot impulse is zero.

1. \(k^2 = \dfrac{L^2}{12} = \dfrac{1}{12}\) m\(^2\). The centre is at \(50\) cm, so \(d = 0.25\) m.

2. \(L_{\text{eq}} = \dfrac{1/12 + 1/16}{1/4} = \dfrac{7/48}{1/4} = \dfrac{7}{12} \approx 0.58\) m. \(T = 2\pi\sqrt{\dfrac{0.583}{10}} \approx 2\pi \times 0.24 \approx 1.5\) s.

Same period means the same equivalent length: \(L_{\text{eq}} = \dfrac{2L}{3} = \dfrac{2 \times 1.5}{3} = 1\) m.

With no pivot impulse the blow \(J\) alone changes the momentum: \(J = m\,v_{\text{cm}} = m\omega\dfrac{L}{2}\). Angular momentum about the pivot: \(Jq = \dfrac{mL^2}{3}\omega\). Dividing the second by the first: \(q = \dfrac{mL^2\omega/3}{m\omega L/2} = \dfrac{2L}{3}\).

The wrist is the pivot. The head, where the blow lands, should be at the centre of percussion, a distance \(\dfrac{I}{md}\) from the wrist. Here \(I\) is the hammer's moment of inertia about the wrist and \(d\) the distance of its centre of mass from the wrist. The hand position sets \(d\) and \(I\); choose it so that this distance equals the wrist-to-head length.

\(T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{1}{10}} = 2\sqrt{\dfrac{\pi^2}{10}} = 2\sqrt{1} = 2\) s.

1. \(L = \dfrac{gT^2}{4\pi^2} = \dfrac{10 \times 4}{40} = 1\) m.

2. \(L \propto T^2\): half the period needs one quarter of the length, \(0.25\) m.

1. The tension is perpendicular to the arc; only the component \(mg\sin\theta\) of the weight acts along the arc, back towards the lowest point: \(F = -mg\sin\theta\).

   For small \(\theta\), \(\sin\theta \approx \theta = \dfrac{x}{L}\), so \(F \approx -\dfrac{mg}{L}x\).

2. \(F = -kx\) with \(k = \dfrac{mg}{L}\), so \(T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{L}{g}}\). The mass cancels.

3. The motion is SHM only while \(\sin\theta \approx \theta\). For large swings the restoring force is smaller than \(-\frac{mg}{L}x\) and the period grows slightly with amplitude.

1. The mass does not appear in \(T = 2\pi\sqrt{L/g}\): still \(T\).

2. \(T \propto \sqrt{L}\): \(2T\).

3. \(T \propto \dfrac{1}{\sqrt{g}}\): \(\sqrt{6}\,T \approx 2.45T\).

1. \(T \propto g^{-1/2}\), so a \(1\) percent fall in \(g\) raises \(T\) by \(\dfrac{1}{2}\) percent. Each swing takes longer, so the clock counts fewer seconds: it loses.

2. \(0.5\) percent of a day \(= 0.005 \times 86400 = 432\) s, about \(7\) minutes.

In the lift the pendulum feels \(g_{\text{eff}} = g + a\) (accelerating up) or \(g - a\) (accelerating down).

1. \(g_{\text{eff}} = 16\): \(T = 2\pi\sqrt{\dfrac{1}{16}} = \dfrac{\pi}{2} \approx 1.57\) s.

2. \(g_{\text{eff}} = 4\): \(T = 2\pi\sqrt{\dfrac{1}{4}} = \pi \approx 3.14\) s.

3. \(g_{\text{eff}} = 0\): no restoring force. The bob stays wherever it is put; there is no oscillation.

1. In the car's frame a pseudo force \(ma\) acts backward. The string settles along the resultant of \(mg\) and \(ma\): \(\tan\theta = \dfrac{a}{g} = 0.75\), so \(\theta \approx 37^\circ\).

2. \(g_{\text{eff}} = \sqrt{g^2 + a^2} = \sqrt{100 + 56.25} = 12.5\) m/s\(^2\).

   \(T = 2\pi\sqrt{\dfrac{1}{12.5}} = 2\sqrt{\dfrac{10}{12.5}} = 2\sqrt{0.8} \approx 1.8\) s.

1. \(F = qE = 10^{-5} \times 10^{5} = 1\) N, equal to the weight \(mg = 1\) N.

   Downward field: the effective weight is \(2\) N, \(g_{\text{eff}} = 20\) m/s\(^2\). \(T = 2\pi\sqrt{\dfrac{1}{20}} = 2\sqrt{\dfrac{10}{20}} = \sqrt{2} \approx 1.41\) s.

2. Upward field: the electric force cancels the weight, \(g_{\text{eff}} = 0\). There is no restoring force and no oscillation.

Buoyancy reduces the effective weight: \(g_{\text{eff}} = g\left(1 - \dfrac{\rho_w}{\rho}\right) = 10\left(1 - \dfrac{1000}{2000}\right) = 5\) m/s\(^2\).

\(T = 2\pi\sqrt{\dfrac{1}{5}} = 2\sqrt{\dfrac{10}{5}} = 2\sqrt{2} \approx 2.8\) s.

1. \(I = \dfrac{1}{3}mL^2\) about the end; \(d = \dfrac{L}{2}\).

2. \(T = 2\pi\sqrt{\dfrac{mL^2/3}{mg(L/2)}} = 2\pi\sqrt{\dfrac{2L}{3g}} = 2\pi\sqrt{\dfrac{2}{30}} = 2\sqrt{\dfrac{10 \times 2}{30}} = 2\sqrt{\dfrac{2}{3}} \approx 1.63\) s.

3. \(2\pi\sqrt{\dfrac{L_{\text{eq}}}{g}} = 2\pi\sqrt{\dfrac{2L}{3g}}\) gives \(L_{\text{eq}} = \dfrac{2L}{3} \approx 0.67\) m. The rod swings like a simple pendulum two thirds as long.

The weight \(mg\) acts at \(G\), a perpendicular distance \(d\sin\theta\) from the pivot: \(\tau = -mgd\sin\theta \approx -mgd\,\theta\) for small \(\theta\).

\(I\alpha = -mgd\,\theta\), so \(\alpha = -\dfrac{mgd}{I}\theta\): angular SHM with \(\omega^2 = \dfrac{mgd}{I}\) and \(T = 2\pi\sqrt{\dfrac{I}{mgd}}\).

1. \(I = I_G + mR^2 = \dfrac{1}{2}mR^2 + mR^2 = \dfrac{3}{2}mR^2\).

2. \(d = R\): \(T = 2\pi\sqrt{\dfrac{3mR^2/2}{mgR}} = 2\pi\sqrt{\dfrac{3R}{2g}} = 2\pi\sqrt{0.03} = 2\sqrt{10 \times 0.03} = 2\sqrt{0.3} \approx 1.1\) s.

\(I = mR^2 + mR^2 = 2mR^2\) about the nail, \(d = R\): \(T = 2\pi\sqrt{\dfrac{2mR^2}{mgR}} = 2\pi\sqrt{\dfrac{2R}{g}} = 2\pi\sqrt{0.04} = 0.4\pi \approx 1.26\) s.

A simple pendulum of length \(2R\) has exactly the same period: the ring's equivalent length is its diameter.

The equivalent length is the length of the simple pendulum with the same period: \(L_{\text{eq}} = \dfrac{I}{md}\).

For \(T = 2\) s: \(L_{\text{eq}} = \dfrac{gT^2}{4\pi^2} = 1\) m, whatever the shape of the body.

\(I = m(k^2 + d^2)\) by the parallel axis theorem, so \(T^2 = 4\pi^2\dfrac{k^2 + d^2}{gd}\).

\(\dfrac{k^2 + d^2}{d} = \dfrac{k^2}{d} + d\). Its derivative \(-\dfrac{k^2}{d^2} + 1\) is zero at \(d = k\), a minimum. Very small \(d\) (pivot near \(G\)) and very large \(d\) both give long periods.

Rod: \(k^2 = \dfrac{L^2}{12}\), so \(d = \dfrac{L}{\sqrt{12}} = \dfrac{1}{3.46} \approx 0.29\) m from the centre.

1. \(T = 2\pi\sqrt{\dfrac{I}{C}} = 2\pi\sqrt{\dfrac{0.01}{0.1}} = 2\pi\sqrt{0.1} = 2\sqrt{10 \times 0.1} = 2\) s.

2. No. The restoring torque is exactly proportional to \(\theta\); no small-angle approximation is needed. So the motion is SHM for any twist within the elastic limit, and the period does not depend on the amplitude.

\(I\alpha = -C\theta\), so \(\alpha = -\dfrac{C}{I}\theta\): angular SHM with \(\omega^2 = \dfrac{C}{I}\) and \(T = 2\pi\sqrt{\dfrac{I}{C}}\).

\(C\) is torque per unit angle: N m per radian, written N m/rad (or simply N m, since the radian has no dimension).

1. \(T \propto \sqrt{I}\): \(2\sqrt{2} \approx 2.8\) s.

2. For a wire \(C \propto \dfrac{1}{\text{length}}\), so doubling the length halves \(C\). \(T \propto \dfrac{1}{\sqrt{C}}\): again \(2\sqrt{2} \approx 2.8\) s.

\(T \propto \sqrt{L}\): a \(0.02\) percent increase in \(L\) gives a \(0.01\) percent increase in \(T\). The clock runs slow and loses.

Loss per day \(= 0.0001 \times 86400 = 8.64\) s.

\(T_1 = 2\) s and \(T_2 = 2\pi\sqrt{\dfrac{4}{10}} = 4\) s.

They are in phase again when both have completed whole numbers of oscillations. After \(4\) s the short pendulum has made \(2\) swings and the long one \(1\).

\(T \propto \sqrt{I}\), so \(\dfrac{I_{\text{disc}} + I_{\text{body}}}{I_{\text{disc}}} = \left(\dfrac{3}{2}\right)^2 = 2.25\).

\(I_{\text{body}} = 1.25 \times 0.01 = 0.0125\) kg m\(^2\). The wire's constant \(C\) cancels out, which makes this a convenient method.

1. \(\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{200}{0.5}} = 20\) rad/s.

2. \(E = \dfrac{1}{2}kA^2 = \dfrac{1}{2} \times 200 \times 0.01 = 1\) J.

3. \(v_{\max} = A\omega = 0.1 \times 20 = 2\) m/s. \(\dfrac{1}{2}mv_{\max}^2 = \dfrac{1}{2} \times 0.5 \times 4 = 1\) J \(= E\): at the mean position all the energy is kinetic.

\(U = \dfrac{1}{2}kx^2 = \dfrac{1}{2} \times 200 \times (0.05)^2 = 0.25\) J, one quarter of \(E\) because \(x = A/2\) and \(U \propto x^2\).

\(K = E - U = 1 - 0.25 = 0.75\) J. \(K : U = 3 : 1\).

\(K = U\) means \(U = \dfrac{E}{2}\): \(\dfrac{1}{2}kx^2 = \dfrac{1}{2} \times \dfrac{1}{2}kA^2\), so \(x^2 = \dfrac{A^2}{2}\) and \(x = \pm\dfrac{A}{\sqrt{2}}\).

For \(A = 10\) cm: \(x \approx \pm 7.1\) cm.

\(U = \dfrac{1}{2}m\omega^2 A^2\sin^2\omega t\). With \(v = A\omega\cos\omega t\), \(K = \dfrac{1}{2}m\omega^2 A^2\cos^2\omega t\).

\(U + K = \dfrac{1}{2}m\omega^2 A^2(\sin^2\omega t + \cos^2\omega t) = \dfrac{1}{2}m\omega^2 A^2\), the same at every instant.

\(v = \omega\sqrt{A^2 - x^2} = \dfrac{A\omega}{2}\) gives \(A^2 - x^2 = \dfrac{A^2}{4}\), so \(x = \dfrac{\sqrt{3}}{2}A = 0.865 \times 10 \approx 8.7\) cm.

\(E = \dfrac{1}{2}kA^2\) depends only on \(k\) and \(A\).

1. \(A \to 2A\): \(E \to 4E\).

2. \(k \to 4k\): \(E \to 4E\).

3. The mass does not appear: \(E\) is unchanged. The heavier block moves more slowly but has the same energy.

\(k = \dfrac{2E}{A^2} = \dfrac{0.8}{0.04} = 20\) N/m. \(\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{20}{0.2}} = 10\) rad/s. \(T = \dfrac{2\pi}{10} = \dfrac{\pi}{5} \approx 0.63\) s.

1. The curves end at \(x = \pm 10\) cm: \(A = 10\) cm. The horizontal line is at \(1\) J: \(E = 1\) J.

2. \(U = \dfrac{1}{2}kx^2\) is proportional to \(x^2\), a parabola opening upward. \(K = E - U = E - \dfrac{1}{2}kx^2\) is the same parabola turned over and lifted to \(E\).

3. They cross where \(K = U = \dfrac{E}{2} = 0.5\) J, at \(x = \pm\dfrac{A}{\sqrt{2}} \approx \pm 7.1\) cm.

1. \(K\) peaks every \(1\) s. The particle passes the mean position twice per period, so the period of the motion is \(2\) s.

2. \(K\) depends on \(v^2\) and \(U\) on \(x^2\). Squaring removes the sign, so the energies repeat every half period, at twice the frequency of the motion.

3. Where the curves cross: \(t = 0.25\) s, \(0.75\) s, \(1.25\) s, and so on, every quarter period.

\(K = \dfrac{1}{2}m\omega^2 A^2\cos^2\omega t = E\cos^2\omega t = \dfrac{E}{2}(1 + \cos 2\omega t)\).

The time-varying part is \(\cos 2\omega t\), of angular frequency \(2\omega\). Over a full period \(\cos 2\omega t\) averages to zero, so the average kinetic energy is \(\dfrac{E}{2}\), and the average potential energy is also \(\dfrac{E}{2}\).

1. At \(P\) and \(Q\) the potential energy equals the total energy, so the kinetic energy is zero: the particle stops and turns back. Beyond them \(U\) would exceed \(E\) and \(K\) would be negative, which is impossible.

2. The gap \(E - U\) is the kinetic energy at that position. It is largest at \(O\).

3. A higher line meets the parabola farther out: \(P\) and \(Q\) move apart, the amplitude grows. The period of SHM does not depend on the amplitude, so it stays the same.

At the extreme: \(K = 0\) and \(U = E\). At the mean position: \(K = E\) and \(U = 0\).

The total energy is never zero while the particle oscillates. It is constant, equal to \(\dfrac{1}{2}kA^2\), and only a stopped particle at the mean position has zero energy.

1. \(E = \dfrac{1}{2}kA^2 = \dfrac{1}{2} \times 100 \times (0.05)^2 = 0.125\) J.

2. \(\dfrac{1}{2}mv_{\max}^2 = 0.125\), so \(v_{\max} = 0.5\) m/s. (Or \(A\omega = 0.05 \times 10\).)

3. \(v = \omega\sqrt{A^2 - x^2} = 10\sqrt{25 - 9}\) cm/s \(= 40\) cm/s \(= 0.4\) m/s.

At the mean position all the energy is kinetic: \(E = \dfrac{1}{2} \times 1 \times 4 = 2\) J.

\(\dfrac{1}{2}kA^2 = 2\) gives \(A^2 = \dfrac{4}{400} = 0.01\), so \(A = 0.1\) m. (Or \(A = \dfrac{v_{\max}}{\omega} = \dfrac{2}{20}\).)

\(E = \dfrac{1}{2}kx^2 + \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 100 \times (0.03)^2 + \dfrac{1}{2} \times 1 \times (0.4)^2 = 0.045 + 0.08 = 0.125\) J.

\(A = \sqrt{\dfrac{2E}{k}} = \sqrt{\dfrac{0.25}{100}} = 0.05\) m \(= 5\) cm. This is the same motion as in Question 13, seen at a different instant.

1. For a small angle \(\theta = \dfrac{A}{L}\), the rise is \(h = L(1 - \cos\theta) \approx L\dfrac{\theta^2}{2} = \dfrac{A^2}{2L}\).

   \(h = \dfrac{0.01}{2} = 5 \times 10^{-3}\) m \(= 5\) mm.

2. \(E = mgh = 0.1 \times 10 \times 5 \times 10^{-3} = 5 \times 10^{-3}\) J. (Check: \(\dfrac{1}{2}m\dfrac{g}{L}A^2\) gives the same.)

   \(\dfrac{1}{2}mv_{\max}^2 = E\): \(v_{\max} = \sqrt{2gh} = \sqrt{0.1} = \dfrac{3.16}{10} \approx 0.32\) m/s.

At the equilibrium position the spring is already stretched by \(x_0 = mg/k\). For a further displacement \(x\) the net force is \(-k(x_0 + x) + mg = -kx\), the same as for a horizontal spring.

Gravity only shifts the equilibrium point. Measured from it, the combined potential energy of spring and gravity is \(\dfrac{1}{2}kx^2\) plus a constant, so the energy equation keeps its usual form.

\(x = A\sin\omega t\), \(U = E\sin^2\omega t\), \(K = E\cos^2\omega t\), so \(U : K = \tan^2\omega t\).

1. \(t = \dfrac{T}{8}\): \(\omega t = 45^\circ\), \(\tan^2 45^\circ = 1\). \(U : K = 1 : 1\).

2. \(t = \dfrac{T}{12}\): \(\omega t = 30^\circ\), \(\tan^2 30^\circ = \dfrac{1}{3}\). \(U : K = 1 : 3\).

Position average: \(\dfrac{1}{A}\displaystyle\int_0^A \dfrac{1}{2}kx^2\,dx = \dfrac{1}{A} \cdot \dfrac{kA^3}{6} = \dfrac{kA^2}{6} = \dfrac{E}{3}\).

The time average is larger, \(\dfrac{E}{2}\), because the particle moves slowly near the extremes and spends more time where \(U\) is large. Averaging equally over positions ignores this.

Only the energy above the minimum of the potential is available for motion: \(K_{\max} = 8 - 2 = 6\) J.

\(\dfrac{1}{2}kA^2 = 6\): \(A^2 = \dfrac{12}{300} = 0.04\), \(A = 0.2\) m.

\(U = \dfrac{1}{2} \times 50 \times (0.2)^2 = 1\) J. \(E = K + U = 3 + 1 = 4\) J.

\(A = \sqrt{\dfrac{2E}{k}} = \sqrt{\dfrac{8}{50}} = \sqrt{0.16} = 0.4\) m.