10 exam-style questions on centripetal force, simple harmonic motion, energy and resonance. Tap Show answer for full working.
A particle moves in a horizontal circle of radius \(0.80\,\mathrm{m}\) at constant speed \(4.0\,\mathrm{m\,s}^{-1}\). Calculate the centripetal acceleration and the time for one complete revolution.
\(a = v^{2}/r = 16/0.80 = 20\,\mathrm{m\,s}^{-2}\).
\(T = 2\pi r/v = 2\pi\times0.80/4.0 = 1.26\,\mathrm{s}\).
A mass of \(0.50\,\mathrm{kg}\) on a string moves in a vertical circle of radius \(1.2\,\mathrm{m}\). At the lowest point its speed is \(6.0\,\mathrm{m\,s}^{-1}\). Calculate the tension in the string at the lowest point. (\(g=9.8\,\mathrm{m\,s}^{-2}\).)
At bottom: \(T - mg = mv^{2}/r\).
\(T = 0.50\times9.8 + 0.50\times36/1.2 = 4.9 + 15 = 19.9\,\mathrm{N}\).
An object in SHM has amplitude \(0.12\,\mathrm{m}\) and frequency \(2.5\,\mathrm{Hz}\). Write the equation for its displacement if it starts from the extreme position at \(t=0\), and calculate its maximum speed.
\(x = 0.12\cos(5\pi t)\) (since \(\omega=2\pi f=5\pi\)).
\(v_{\max} = \omega A = 5\pi\times0.12 = 1.88\,\mathrm{m\,s}^{-1}\).
A car of mass \(1200\,\mathrm{kg}\) travels over a convex bridge of radius \(50\,\mathrm{m}\) at \(15\,\mathrm{m\,s}^{-1}\). Calculate the normal reaction on the car at the highest point. (\(g=9.8\,\mathrm{m\,s}^{-2}\).)
At top: \(mg - N = mv^{2}/r\).
\(N = 1200\times9.8 - 1200\times225/50 = 11760 - 5400 = 6360\,\mathrm{N}\).
A simple pendulum of length \(1.5\,\mathrm{m}\) has a period of \(2.45\,\mathrm{s}\) on Earth. Calculate the acceleration due to gravity at that location. What would the period be on the Moon where \(g=1.6\,\mathrm{m\,s}^{-2}\)?
\(T = 2\pi\sqrt{L/g} \Rightarrow g = 4\pi^{2}L/T^{2} = 4\pi^{2}\times1.5/(2.45)^{2} = 9.86\,\mathrm{m\,s}^{-2}\).
On Moon: \(T = 2\pi\sqrt{1.5/1.6} = 6.08\,\mathrm{s}\).
A mass–spring system has mass \(0.40\,\mathrm{kg}\) and spring constant \(160\,\mathrm{N\,m}^{-1}\). It oscillates with amplitude \(0.050\,\mathrm{m}\). Calculate the maximum kinetic energy and the speed when the displacement is \(0.030\,\mathrm{m}\).
\(\omega = \sqrt{k/m} = \sqrt{400} = 20\,\mathrm{rad\,s}^{-1}\).
\(E_{\mathrm{tot}} = \frac12 kA^{2} = 0.5\times160\times0.0025 = 0.20\,\mathrm{J}\) (maximum KE).
\(v = \omega\sqrt{A^{2}-x^{2}} = 20\sqrt{0.0025-0.0009} = 20\sqrt{0.0016} = 0.80\,\mathrm{m\,s}^{-1}\).
Explain what is meant by resonance. Describe how the amplitude of a driven oscillator varies with driving frequency, and state the condition for maximum amplitude when damping is light.
Resonance: large amplitude oscillations when a system is driven at (or near) its natural frequency.
Amplitude is small at low and high frequencies, rises to a peak near the natural frequency. For light damping the peak occurs at a frequency very close to the natural frequency \(\omega_{0}\).
A particle moves in a horizontal circle of radius \(r\) with constant speed \(v\). Derive an expression for the centripetal force in terms of mass \(m\), \(v\) and \(r\). A satellite of mass \(250\,\mathrm{kg}\) orbits the Earth at a height of \(400\,\mathrm{km}\) above the surface. Radius of Earth \(= 6400\,\mathrm{km}\), \(g=9.8\,\mathrm{m\,s}^{-2}\) at surface. Calculate the orbital speed assuming \(g\) varies as \(1/r^{2}\).
Acceleration \(a = v^{2}/r\) (from \(a = \omega^{2}r\) or geometry of velocity change). Force \(F = ma = mv^{2}/r\).
Orbital radius \(r = 6800\,\mathrm{km} = 6.80\times10^{6}\,\mathrm{m}\).
\(g(r) = g_{0}(R/r)^{2} = 9.8\times(6400/6800)^{2} = 8.68\,\mathrm{m\,s}^{-2}\).
\(v = \sqrt{gr} = \sqrt{8.68\times6.80\times10^{6}} = 7680\,\mathrm{m\,s}^{-1}\).
The displacement of a particle in SHM is \(x = 0.08\sin(10t + \pi/4)\) (SI units). Determine the amplitude, period, maximum acceleration and the displacement at \(t=0\).
\(A = 0.08\,\mathrm{m}\); \(\omega = 10\,\mathrm{rad\,s}^{-1}\); \(T = 2\pi/\omega = 0.628\,\mathrm{s}\).
\(a_{\max} = \omega^{2}A = 100\times0.08 = 8.0\,\mathrm{m\,s}^{-2}\).
At \(t=0\): \(x = 0.08\sin(\pi/4) = 0.08\times\sqrt{2}/2 = 0.0566\,\mathrm{m}\).
A mass \(m\) is attached to a light spring of stiffness \(k\) on a smooth horizontal surface. The mass is pulled a distance \(A\) from equilibrium and released. Show that the motion is simple harmonic and find expressions for the period and the maximum kinetic energy.
Restoring force \(F = -kx\). Acceleration \(a = F/m = -(k/m)x\). This is of the form \(a = -\omega^{2}x\) with \(\omega^{2} = k/m\), so SHM.
\(T = 2\pi\sqrt{m/k}\).
Maximum KE \(=\) total energy \(= \frac12 kA^{2}\).