Energy in oscillatory motion
A mass-spring oscillator neither gains nor loses energy while it oscillates without friction: it only transforms it, over and over, between two forms. At the extremes all the energy is elastic potential energy, stored in the stretched or compressed spring; when passing through the equilibrium position all of it is kinetic, belonging to the body moving at its maximum speed. Understanding this back-and-forth is understanding why the amplitude of a free oscillation is constant.
Elastic potential energy and kinetic energy
When you pull back the plunger of a pinball machine in an arcade in Salthill, pulling it twice as far gives the ball much more energy, in fact four times as much: how exactly does the stored energy depend on the distance it is deformed?
By Hooke's law the force needed to stretch the spring is $F=ks$ (the spring pulls back with $F=-ks$). The work done in stretching it is the area under the $F$–$s$ graph, a triangle of area $\tfrac12 ks^2$. That work is stored as $E_P=\tfrac12 ks^2$, which is why it grows with the square of the displacement. You can verify Hooke's law in the lab: hang increasing loads on a spring, measure the extension each time and plot force against extension; the points lie on a straight line whose gradient is $k$.
A mass attached to a spring oscillates horizontally without friction between positions $x = -A$ and $x = +A$. The figure shows the oscillator with these positions…
A mass attached to a spring oscillates horizontally without friction between positions $x = -A$ and $x = +A$. The figure shows the oscillator with these positions marked.
- a) At which position or positions is the elastic potential energy maximum?
- b) At which position is the kinetic energy maximum?
- c) At which positions is the kinetic energy zero? Justify your answers using formulas $E_p = \frac{1}{2} \times k \times x^2$ and $E_c = \frac{1}{2} \times m \times v^2$.
Mechanical energy: it is conserved
A sensor records a mass-spring oscillator: at the extremes ($x = \pm A$) the body stops completely, whereas on crossing the centre ($x = 0$) it reaches its maximum speed and the spring exerts no pull. Does the total energy of the system change during this continuous exchange, or does it remain strictly constant?
A trolley of $m = 0.5\ \text{kg}$ attached to a spring passes through the equilibrium position with a velocity of $2\ \text{m/s}$. a) Calculate its kinetic energy at…
A trolley of $m = 0.5\ \text{kg}$ attached to a spring passes through the equilibrium position with a velocity of $2\ \text{m/s}$.
- a) Calculate its kinetic energy at that point.
- b) What is its elastic potential energy at that point? And its mechanical energy? Justify your answer.
How Ep and Ec are shared out according to position
The same principle in the pendulum, and real energy loss
On a trampoline both kinds of potential energy are involved: at the lowest point the mat stores the most elastic energy, at the highest point the jumper has the most gravitational energy, and the kinetic energy is greatest in between.
For a pendulum at small angles ($\theta \le 10^\circ$) the period is $T = 2\pi\sqrt{l/g}$. To measure it, time 10 complete swings with a stopwatch and divide by 10; the uncertainty of $T$ is the uncertainty of the total time divided by 10. Repeating the measurement and comparing the energy at the start and after many swings shows how much is lost to friction.
A gymnast jumps on a trampoline and oscillates vertically, without losing contact with the mat, between a highest point and a lowest point. a) At which point of the…
A gymnast jumps on a trampoline and oscillates vertically, without losing contact with the mat, between a highest point and a lowest point.
- a) At which point of the oscillation is her gravitational potential energy maximum?
- b) At which point is the elastic potential energy stored by the mat maximum?
- c) Where is her kinetic energy maximum? Justify all three answers by energy exchange.
The steel ball of a wrecking crane hangs from a cable of $l = 2\ \text{m}$ and is released from rest when the cable forms an angle of $60^\circ$ with the vertical (take…
The steel ball of a wrecking crane hangs from a cable of $l = 2\ \text{m}$ and is released from rest when the cable forms an angle of $60^\circ$ with the vertical (take $g = 9.8\ \text{m/s}^2$). The height it descends relative to the lowest point is $h = l \times (1 - \cos 60^\circ)$. The situation is shown in the figure.
- a) Calculate that height $h$.
- b) Apply conservation of energy, $m \times g \times h = \frac{1}{2} \times m \times v^2$, and calculate the speed of the ball when passing through the lowest point.
- c) Does this speed depend on the mass of the ball? Justify your answer.
A real oscillator with friction loses $10\ \%$ of its mechanical energy in each cycle. It starts with an initial energy $E_0 = 2\ \text{J}$. The figure shows how the…
A real oscillator with friction loses $10\ \%$ of its mechanical energy in each cycle. It starts with an initial energy $E_0 = 2\ \text{J}$. The figure shows how the amplitude of oscillation decreases with time.
- a) Calculate the remaining mechanical energy after $5$ cycles.
- b) Given that $E_m = \frac{1}{2} \times k \times A^2$, that is, energy is proportional to the square of the amplitude, calculate the factor by which the amplitude has been reduced after those $5$ cycles.
- c) What has the lost energy been transformed into? Explain why the period of oscillation, however, hardly changes even though the amplitude decreases (hint: on which quantities does $T$ depend?).
A laboratory demonstration pendulum has a length $l = 1\ \text{m}$ and is released from rest forming $60^\circ$ with the vertical (take $g = 9.8\ \text{m/s}^2$). Recall…
A laboratory demonstration pendulum has a length $l = 1\ \text{m}$ and is released from rest forming $60^\circ$ with the vertical (take $g = 9.8\ \text{m/s}^2$). Recall that the height above the lowest point is $h = l \times (1 - \cos\theta)$. The figure shows the pendulum in the three positions considered.
- a) Calculate the maximum speed it reaches when passing through the lowest point.
- b) Calculate its speed when the string forms $30^\circ$ with the vertical, applying conservation of energy between the points at $60^\circ$ and $30^\circ$.
- c) Compare the two speeds: why is the speed at $30^\circ$ more than half of the maximum speed even though the angle is exactly half?
Back to the start: The amplitude stays constant precisely because mechanical energy is conserved ($E_m = \frac{1}{2}\cdot k\cdot A^2 = \text{const}$); since the stiffness $k$ of the spring is fixed, the value of $A$ cannot change from one swing to the next unless friction comes into play.