Energy in oscillatory motion

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.

1. a) At which position or positions is the elastic potential energy maximum?
2. b) At which position is the kinetic energy maximum?
3. 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 $E_c = \frac{1}{2} \times m \times v^2$.

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, $E_p = \frac{1}{2} \times k \times x^2$.

The mechanical energy of a mass-spring oscillator is $E_m = \frac{1}{2} \times k \times A^2$, 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 $2\ \text{m/s}$.

1. a) Calculate its kinetic energy at that point.
2. 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 with amplitude $A = 0.08\ \text{m}$.

1. a) Calculate the mechanical energy of the vibration.
2. b) All this energy is kinetic when passing through equilibrium: calculate the maximum speed of the diaphragm.
3. c) Check that it matches the value given by the expression $v_{max} = A \times \sqrt{\frac{k}{m}}$.

An oscillator of $m = 0.2\ \text{kg}$ and $k = 50\ \text{N/m}$ oscillates with amplitude $A = 0.2\ \text{m}$.

1. a) Calculate its mechanical energy.
2. b) Calculate the potential energy, kinetic energy and speed when the displacement is $x = 0.16\ \text{m}$.
3. c) Show, using $E_c = \frac{1}{2} \times k \times (A^2 - x^2)$, that the speed is half of the maximum speed when $x = A \times \frac{\sqrt{3}}{2}$, and calculate that position.

A mass-spring oscillator has a constant mechanical energy $E_m$.

1. a) Describe how the elastic potential energy $E_p = \frac{1}{2} \times k \times x^2$ varies with position: what shape does its graph against $x$ have between $-A$ and $+A$?
2. b) The kinetic energy at each point is $E_c = E_m - E_p$. Describe how $E_c$ varies with position.
3. 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 shows how the oscillator's energy is distributed as a function of position.

1. a) Calculate the mechanical energy of the oscillator, $E_m = \frac{1}{2} \times k \times A^2$.
2. b) Calculate the elastic potential energy when the displacement is $x = 0.06\ \text{m}$.
3. 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 $A = 0.1\ \text{m}$.

1. a) Calculate the mechanical energy and the elastic potential energy at points $x = 0$, $x = 0.05\ \text{m}$ and $x = 0.1\ \text{m}$.
2. b) Calculate the kinetic energy at these same three points.
3. 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 $E_c$.
4. 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.

A gymnast jumps on a trampoline and oscillates vertically, without losing contact with the mat, between a highest point and a lowest point.

1. a) At which point of the oscillation is her gravitational potential energy maximum?
2. b) At which point is the elastic potential energy stored by the mat maximum?
3. 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 $g = 9.8\ \text{m/s}^2$). The height it descends relative to the lowest point is $h = l \times (1 - \cos 60^\circ)$. The situation is shown in the figure.

1. a) Calculate that height $h$.
2. 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.
3. c) Does this speed depend on the mass of the ball? Justify your answer.

A toy launches a ball of $m = 0.05\ \text{kg}$ vertically using a spring of constant $k = 500\ \text{N/m}$ that is compressed by $x = 0.1\ \text{m}$ before release (take $g = 9.8\ \text{m/s}^2$). The launch is shown in the figure.

1. a) Calculate the elastic potential energy stored in the compressed spring.
2. b) Assuming that all this energy is converted into kinetic energy of the ball, calculate its launch speed.
3. c) Calculate the maximum height the ball will reach, applying conservation of energy again.
4. d) The actual toy reaches a lower height than calculated. State at least two energetic reasons that explain this.

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.

1. a) Calculate the remaining mechanical energy after $5$ cycles.
2. 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.
3. 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 that the height above the lowest point is $h = l \times (1 - \cos\theta)$. The figure shows the pendulum in the three positions considered.

1. a) Calculate the maximum speed it reaches when passing through the lowest point.
2. b) Calculate its speed when the string forms $30^\circ$ with the vertical, applying conservation of energy between the points at $60^\circ$ and $30^\circ$.
3. 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?