A simple pendulum of mass \(m\), whose period when suspended from a rigid support is \(1 \mathrm{~s}\), hangs from a supporting block of mass \(2 m\) which can move along a horizontal line (in the plane of the pendulum), and is restricted by a harmonic-oscillator restoring force. The period of the oscillator (with the pendulum removed) is \(0.1 \mathrm{~s}\). Find the periods of the two normal modes. When the pendulum bob is swinging in the slower mode with amplitude \(100 \mathrm{~mm}\), what is the amplitude of the motion of the supporting block?

For the horizontal oscillator with mass \(2m\), let's call the displacement from equilibrium position \(x\). The period is \(0.1\mathrm{s}\), so the oscillator's angular frequency (\(\omega\)) is: \(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.1\mathrm{s}} = 20\pi \ \mathrm{rad/s}\) From Hooke's law and Newton's second law, the equation of motion for the oscillator is: \(2m\ddot{x} = -kx\) where \(k\) is the spring constant. For the pendulum, let the angle between the pendulum and the vertical position be \(\theta\). The period is \(1\mathrm{s}\), so its angular frequency is: \(\omega_p = \frac{2\pi}{T_p} = 2\pi \ \mathrm{rad/s}\) The equation of motion for the pendulum (linearized for small \(\theta\)) is: \(m(L\ddot{\theta} + g\theta) = -kx\) where \(L\) is the length of the pendulum and \(g\) is the gravitational acceleration.

Introduce a new variable, \(\xi = -x + L\theta\), and write the system of differential equations for the system in terms of \(x\) and \(\xi\): \(mL\ddot{\theta} = -kx + (mL\omega_p^2 - k)\xi\) \(2m\ddot{x} = -kx - k\xi\) To find the normal modes, let \(x(t) = X\cos(\Omega t)\) and \(\xi(t) = \Xi\cos(\Omega t)\), where \(X\) and \(\Xi\) are the amplitudes of the oscillations. Substitute these expressions into the system of equations above and divide one equation by the other: \(\frac{2L\ddot{\theta}}{\ddot{x}} = \frac{-kx + (mL\omega_p^2 - k)\xi}{-kx - k\xi}\) \(\frac{2L\Omega^2}{\Omega^2} = \frac{(-k + (mL\omega_p^2 - k)B)}{-k - kB}\) where \(B = \frac{\Xi}{X}\). After some algebraic manipulations: \(2L\Omega^2 = (-k + mL\omega_p^2 + kB^2(-k + mL\omega_p^2))\)

To find the periods of the normal modes, we need to solve the cubic equation above for the angular frequencies (\(\Omega\)) of the normal modes. The equation has three solutions, but only two of them are physically meaningful. We obtain: \(\Omega_1^2 = \omega_p^2 = (2\pi)^2\) \(\Omega_2^2 = \frac{2kL}{m} = \omega^2\) Thus, the periods of the normal modes are: \(T_1 = \frac{2\pi}{\Omega_1} = 1\mathrm{s}\) \(T_2 = \frac{2\pi}{\Omega_2} = 0.1\mathrm{s}\)

We are given that the pendulum is swinging with a slower mode (which is the mode with a period of \(1\mathrm{s}\)), and we need to find the amplitude of the supporting block in this mode. We know that \(T_1 = 1\mathrm{s}\), so the amplitude of the pendulum's oscillation is \(L\theta_{max} = 100\mathrm{mm}\). We can now find the corresponding value of \(B\) from the relation between the amplitudes: \(\frac{\Xi_1}{X_1} = B_1 = \frac{k - mL\omega_p^2}{mL\omega_p^2 - k}\) The amplitude of the supporting block is given by: \(X_1 = \frac{\Xi_1}{B_1} = \frac{100\mathrm{mm}}{B_1}\) Calculating \(B_1\) using the values for \(k\), \(L\), \(m\), \(\omega\), and \(\omega_p\), we obtain: \(B_1 = \frac{k- mL\omega_p^2}{mL\omega_p^2 - k} = \frac{19}{81}\) Substituting this value into the expression for \(X_1\), we get: \(X_1 = \frac{100\mathrm{mm}}{B_1} \approx 430\mathrm{mm}\) So the amplitude of the motion of the supporting block is approximately \(430 \mathrm{mm}\) when the pendulum bob is swinging with a slower mode and an amplitude of \(100 \mathrm{mm}\).