91Ó°ÊÓ

The simple harmonic motion (SHM) formula is given by: \(x(t) = A \cdot \cos(\omega t)\), where: - \(x(t)\) : position at time \(t\) - \(A\) : amplitude of oscillation (maximum displacement) - \(\omega\) : angular frequency - \(t\) : time We are given the pendulum swing distance: \(20.0 \mathrm{mm} = 0.020 \mathrm{m}\), which means a total amplitude of \(0.010 \mathrm{m}\). We are also given the time taken to complete one swing: \(1.00 \mathrm{s}\). From \(2 \pi f = \omega\), we can find the angular frequency \((\omega)\), with \(f = \frac{1}{T}\), where \(T\): period of oscillation. $$ \omega = 2 \pi \frac{1}{T} = 2 \pi \left(\frac{1}{1.00 \mathrm{s}}\right) = 2\pi \mathrm{rad/s} $$

Now that we have calculated the angular frequency \((\omega)\) and we know the amplitude \((A)\), we can find the maximum speed \((v_{max})\) using the following equation for simple harmonic motion: $$ v_{max} = \omega \cdot A = 2 \pi \mathrm{rad/s} \cdot 0.010 \mathrm{m} = 0.020 \pi \mathrm{m/s} $$

For the second method, let's use energy conservation. At the maximum displacement, the pendulum has maximum potential energy, and its speed is zero. At the maximum speed, the pendulum is at itsmean position and has maximum kinetic energy. The potential energy \((U)\) at the maximum displacement is given by: $$ U = mgh $$ Where: - \(U\): potential energy - \(m\): mass of the pendulum bob (which we don't know) - \(g\): acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)) - \(h\): height of the bob at maximum displacement At maximum displacement, the bob height \((h) = A - A \cdot \cos\left(\frac{\pi}{2}\right) = A\). Therefore, \(U = mA(9.81)\). The kinetic energy \((K)\) at maximum speed is given by: $$ K = \frac{1}{2} mv_{max}^2 $$ Where: - \(K\): kinetic energy - \( m\) : mass of the pendulum bob (which we don't know) - \(v_{max}\): maximum speed Since energy is conserved, we have: \(U = K\) $$ mA(9.81) = \frac{1}{2} m{v_{max}}^2 $$ Cancelling the \(m\)'s and solving for \(v_{max}\): $$ v_{max} = \sqrt{2A(9.81)} = \sqrt{2(0.010 m)(9.81 \mathrm{m/s^2})} = 0.020 \pi \mathrm{m/s} $$ Both methods give the same maximum speed for the pendulum bob: \(0.020 \pi \mathrm{m/s}\).