The Workbook

1. Compare with \(x = A\sin\omega t\): \(A = 0.05\) m, \(\omega = 10\pi\) rad/s. \(T = \dfrac{2\pi}{\omega} = 0.2\) s, \(f = \dfrac{1}{T} = 5\) Hz.

2. \(v_{\max} = A\omega = 0.05 \times 10\pi = 0.5\pi \approx 1.57\) m/s. \(a_{\max} = A\omega^2 = 0.05 \times 100\pi^2 = 5\pi^2 \approx 49\) m/s\(^2\).

1. \(A = 4\) cm; \(\omega = 2\pi\) rad/s gives \(T = 1\) s; initial phase \(\dfrac{\pi}{3}\).

2. \(x(0) = 4\cos\dfrac{\pi}{3} = 2\) cm.

3. \(v = -8\pi\sin\left(2\pi t + \dfrac{\pi}{3}\right)\). At \(t = 0\), \(v = -8\pi\sin\dfrac{\pi}{3} < 0\): the particle moves towards \(-x\).

1. The force is proportional to the displacement and directed towards \(O\) (the minus sign). Such a force gives SHM.

   \(a = \dfrac{F}{m} = -100x = -\omega^2 x\), so \(\omega^2 = 100\) and \(\omega = 10\) rad/s.

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

\(v = \omega\sqrt{A^2 - x^2} = 5\sqrt{100 - 36} = 5 \times 8 = 40\) cm/s.

\(a = -\omega^2 x = -25 \times 6 = -150\) cm/s\(^2\): \(150\) cm/s\(^2\) directed towards the mean position.

1. Speed is largest at the mean position (\(A\omega\)) and zero at the extremes.

2. Acceleration is largest at the extremes (\(A\omega^2\)) and zero at the mean position.

3. The restoring force \(-kx\) is largest at the extremes and zero at the mean position, like the acceleration.

\(v^2 = \omega^2(A^2 - x^2)\): \(64 = \omega^2(A^2 - 9)\) and \(36 = \omega^2(A^2 - 16)\).

Subtract: \(28 = 7\omega^2\), so \(\omega = 2\) rad/s. Then \(A^2 - 9 = 16\), \(A = 5\) cm.

\(\omega = \dfrac{2\pi}{T} = \pi\) rad/s.

1. Starting at the positive extreme: \(x = 3\cos\pi t\) (cm).

2. Starting at the mean position and moving towards \(-x\): \(x = -3\sin\pi t\) (cm). Check: \(v = -3\pi\cos\pi t\) is negative at \(t = 0\).

1. The curve reaches \(\pm 2\) cm: \(A = 2\) cm. It repeats every \(4\) s: \(T = 4\) s, so \(\omega = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\) rad/s.

   It starts at zero and rises: \(x = 2\sin\dfrac{\pi t}{2}\) (cm).

2. \(v_{\max} = A\omega = 2 \times \dfrac{\pi}{2} = \pi\) cm/s.

3. The particle is at rest at the extremes, where the graph peaks: \(t = 1\) s and \(t = 3\) s.

1. \(v = A\omega\cos\omega t = A\omega\sin\left(\omega t + \dfrac{\pi}{2}\right)\): \(v\) leads \(x\) by \(\dfrac{\pi}{2}\).

   \(a = -A\omega^2\sin\omega t = A\omega^2\sin(\omega t + \pi)\): \(a\) is in opposite phase to \(x\), a phase difference of \(\pi\).

2. At \(x = A\) (\(\omega t = \pi/2\)): \(v = 0\) and \(a = -\omega^2 A\), the largest acceleration, pointing back towards \(O\).

1. The graph is a straight line through the origin with negative slope, so \(a = -(\text{constant})x\). This is the condition for SHM.

2. Slope \(= \dfrac{-100}{4} = -25\) s\(^{-2}\), so \(\omega^2 = 25\) and \(\omega = 5\) rad/s. \(T = \dfrac{2\pi}{5} \approx 1.26\) s.

3. The line ends at \(x = \pm 4\) cm, the largest displacement: \(A = 4\) cm.

1. \(v = 0\) at \(x = \pm 5\) cm: \(A = 5\) cm. \(x = 0\) at \(v = \pm 10\) cm/s: \(v_{\max} = 10\) cm/s. \(\omega = \dfrac{v_{\max}}{A} = 2\) rad/s.

2. \(v^2 = \omega^2(A^2 - x^2)\) gives \(\dfrac{x^2}{A^2} + \dfrac{v^2}{A^2\omega^2} = 1\), the equation of an ellipse with semi-axes \(A\) along \(x\) and \(A\omega\) along \(v\).

1. \(\theta = \omega t\), and \(OQ = A\sin\theta\). So \(x = A\sin\omega t\).

2. \(P\) has centripetal acceleration \(\omega^2 A\) towards \(O\). Its component along the vertical diameter is \(-\omega^2 A\sin\theta = -\omega^2 x\), and this is the acceleration of \(Q\).

   \(a = -\omega^2 x\) is the defining equation of SHM.

3. The radius of the circle is the amplitude, and the angular speed of \(P\) is the angular frequency \(\omega\) of the SHM. The angle \(\theta\) is the phase.

1. \(x = A\sin\theta = \dfrac{A}{2}\) needs \(\theta = 30^\circ\). \(P\) turns \(360^\circ\) in \(T\), so \(30^\circ\) takes \(\dfrac{T}{12}\).

2. From \(\theta = 30^\circ\) to \(90^\circ\) is \(60^\circ\): a time \(\dfrac{T}{6}\), twice as long.

The particle slows down as it nears the extreme, so the outer half of the amplitude takes longer than the inner half.

1. \(\sin\theta = \dfrac{1}{\sqrt{2}}\) at \(\theta = 45^\circ\). From \(45^\circ\) to \(90^\circ\) is \(45^\circ\): \(\dfrac{T}{8}\).

2. From \(\theta = -30^\circ\) to \(+30^\circ\) is \(60^\circ\): \(\dfrac{T}{6}\).

\(x = 3\) cm is \(\dfrac{A}{2}\): time \(= \dfrac{T}{12} = 1\) s. From \(\dfrac{A}{2}\) to \(A\): \(\dfrac{T}{6} = 2\) s.

1. \(x_2 = A\cos\omega t = A\sin\left(\omega t + \dfrac{\pi}{2}\right)\): \(x_2\) leads \(x_1\) by \(\dfrac{\pi}{2}\). It reaches its maximum a quarter period earlier.

2. \(x_3 = -A\sin\omega t = -x_1\). The two are always at opposite displacements: opposite phase, a difference of \(\pi\).

\(|x| > \dfrac{A}{2}\) means \(|\sin\theta| > \dfrac{1}{2}\): \(\theta\) between \(30^\circ\) and \(150^\circ\), or between \(210^\circ\) and \(330^\circ\).

That is \(120^\circ + 120^\circ = 240^\circ\) out of \(360^\circ\), a fraction \(\dfrac{2}{3}\) of the period. The particle moves slowly near the extremes and lingers there.

\(Q\) at the mean position moving towards \(+x\) has phase \(0\) (\(x = A\sin\omega t\)). \(P\) at \(x = +A\) has phase \(\dfrac{\pi}{2}\).

The phase difference is \(\dfrac{\pi}{2}\), and \(P\) leads: \(Q\) will be where \(P\) is a quarter period later.

1. In one period the particle covers \(4A = 20\) cm. Average speed \(= \dfrac{20}{2} = 10\) cm/s.

2. \(v_{\max} = A\omega = 5 \times \pi = 5\pi \approx 15.7\) cm/s. In general \(\dfrac{\text{average}}{v_{\max}} = \dfrac{4A/T}{2\pi A/T} = \dfrac{2}{\pi}\).

\(f = \dfrac{600}{60} = 10\) Hz, \(\omega = 20\pi\) rad/s, \(A = 0.05\) m.

\(v_{\max} = A\omega = 0.05 \times 20\pi = \pi \approx 3.1\) m/s. \(a_{\max} = A\omega^2 = 0.05 \times 400\pi^2 = 20\pi^2 = 200\) m/s\(^2\), twenty times \(g\).

The block leaves when the platform accelerates downward faster than \(g\). This first happens at the top of the motion, where the downward acceleration is \(A\omega^2\).

\(A\omega^2 = g\): \(\omega^2 = \dfrac{10}{0.1} = 100\), \(\omega = 10\) rad/s, \(f = \dfrac{\omega}{2\pi} = \dfrac{5}{\pi} \approx 1.6\) Hz. The block leaves at the highest point.

\(\omega = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\) rad/s and \(x = 6\sin\dfrac{\pi t}{2}\). At \(t = \dfrac{1}{3}\) s the phase is \(\dfrac{\pi}{6}\).

\(x = 6\sin\dfrac{\pi}{6} = 3\) cm. \(v = A\omega\cos\dfrac{\pi}{6} = 6 \times \dfrac{\pi}{2} \times \dfrac{1.73}{2} \approx 8.2\) cm/s, towards \(+x\).