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 a catapult, you notice that stretching it twice as far takes much more effort and that the projectile is launched with four times the energy: how exactly does the stored energy depend on the distance it is deformed?
The work done by the elastic force when deforming the spring by a distance $x$ is exactly the energy that ends up stored in it: that is why $E_p$ grows with the square of $x$, just like the area under the line $F=k\cdot x$ of Hooke's law.
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 $E_p = \frac{1}{2} \cdot k \cdot x^2$ and $E_c = \frac{1}{2} \cdot m \cdot v^2$.
The pulleys and bow of a target archery bow together behave like a spring with an equivalent constant of $k = 300\ \text{N/m}$. When drawn, the string is displaced by…
The pulleys and bow of a target archery bow together behave like a spring with an equivalent constant of $k = 300\ \text{N/m}$. When drawn, the string is displaced by $x = 0,1\ \text{m}$. Calculate the stored elastic potential energy, $E_p = \frac{1}{2} \cdot k \cdot x^2$.
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?
The mechanical energy of a mass-spring oscillator is $E_m = \frac{1}{2} \cdot k \cdot A^2$, where $A$ is the amplitude. In the laboratory, an oscillator has…
The mechanical energy of a mass-spring oscillator is $E_m = \frac{1}{2} \cdot k \cdot A^2$, where $A$ is the amplitude. In the laboratory, an oscillator has $k = 200\ \text{N/m}$ and oscillates with an amplitude of $A = 0,05\ \text{m}$. Calculate its mechanical energy.
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.
The diaphragm of a loudspeaker can be modelled as a mass-spring oscillator of mass $m = 0,4\ \text{kg}$ and effective spring constant $k = 250\ \text{N/m}$, vibrating…
The diaphragm of a loudspeaker can be modelled as a mass-spring oscillator of mass $m = 0,4\ \text{kg}$ and effective spring constant $k = 250\ \text{N/m}$, vibrating with amplitude $A = 0,08\ \text{m}$.
- a) Calculate the mechanical energy of the vibration.
- b) All this energy is kinetic when passing through equilibrium: calculate the maximum speed of the diaphragm.
- c) Check that it matches the value given by the expression $v_{max} = A \cdot \sqrt{\frac{k}{m}}$.
How Ep and Ec are shared out according to position
A mass-spring oscillator has a constant mechanical energy $E_m$. a) Describe how the elastic potential energy $E_p = \frac{1}{2} \cdot k \cdot x^2$ varies with position:…
A mass-spring oscillator has a constant mechanical energy $E_m$.
- a) Describe how the elastic potential energy $E_p = \frac{1}{2} \cdot k \cdot x^2$ varies with position: what shape does its graph against $x$ have between $-A$ and $+A$?
- b) The kinetic energy at each point is $E_c = E_m - E_p$. Describe how $E_c$ varies with position.
- c) What does the point where the graphs of $E_c$ and $E_p$ intersect represent? Draw the three graphs ($E_p$, $E_c$ and $E_m$) against position in your notebook.
You can do it on paper and send a photo with your answer.
A trolley of $m = 0,5\ \text{kg}$ oscillates on a frictionless table attached to a spring of $k = 400\ \text{N/m}$, with an amplitude of $A = 0,1\ \text{m}$. The figure…
A trolley of $m = 0,5\ \text{kg}$ oscillates on a frictionless table attached to a spring of $k = 400\ \text{N/m}$, with an amplitude of $A = 0,1\ \text{m}$. The figure shows how the oscillator's energy is distributed as a function of position.
- a) Calculate the mechanical energy of the oscillator, $E_m = \frac{1}{2} \cdot k \cdot A^2$.
- b) Calculate the elastic potential energy when the displacement is $x = 0,06\ \text{m}$.
- c) Calculate the kinetic energy at that same point and, from it, the speed of the trolley.
A mass-spring oscillator has $k = 100\ \text{N/m}$ and oscillates with amplitude $A = 0,1\ \text{m}$. a) Calculate the mechanical energy and the elastic potential energy…
A mass-spring oscillator has $k = 100\ \text{N/m}$ and oscillates with amplitude $A = 0,1\ \text{m}$.
- a) Calculate the mechanical energy and the elastic potential energy at points $x = 0$, $x = 0,05\ \text{m}$ and $x = 0,1\ \text{m}$.
- b) Calculate the kinetic energy at these same three points.
- c) Using these values, draw the graphs of $E_p(x)$, $E_c(x)$ and $E_m(x)$ between $-A$ and $+A$ in your notebook and describe them: shape of each curve, intercepts with the axes and intersection point between $E_p$ and $E_c$.
- d) Calculate the exact position where the graphs of $E_p$ and $E_c$ intersect.
You can do it on paper and send a photo with your answer.
The same principle in the pendulum, and real energy loss
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 $h = l \cdot (1 - \cos 60^\circ)$. The situation is shown in the figure.
- a) Calculate that height $h$.
- b) Apply conservation of energy, $m \cdot g \cdot h = \frac{1}{2} \cdot m \cdot v^2$, 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.
During the spin cycle, the drum of a washing machine vibrates with a mechanical energy of $1,44\ \text{J}$. The shock absorbers supporting it behave like a spring with…
During the spin cycle, the drum of a washing machine vibrates with a mechanical energy of $1,44\ \text{J}$. The shock absorbers supporting it behave like a spring with an effective constant of $k = 200\ \text{N/m}$.
- a) Rearrange formula $E_m = \frac{1}{2} \cdot k \cdot A^2$ for amplitude.
- b) Calculate the amplitude of vibration, in metres, and express it in centimetres as well.
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 $E_m = \frac{1}{2} \cdot k \cdot A^2$, 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?).
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.