AP Physics 1 Exam Question - Periodic Motion

A simple pendulum consists of a mass hanging from a string of length $L$. The mass is pulled to one side and released, causing it to swing back and forth. The period $T$ of the pendulum is defined as the time it takes for the mass to complete one full swing. Consider a simple pendulum with a length $L = 1.5$ m and a mass $m = 0.2$ kg.

(a) Calculate the period $T$ of this simple pendulum.

(b) If the mass is pulled aside with an angle $\theta$ of $30^\circ$ from the vertical and released from rest, calculate the maximum speed $v_{\text{max}}$ of the mass as it swings through its equilibrium position.

(c) Using energy considerations, calculate the height $h$ above the equilibrium position that the mass reaches during its swing.

(d) Calculate the tension $T_{\text{max}}$ in the string when the mass is at its maximum height.

Give all numerical answers with the appropriate unit.

Solution

(a) To find the period $T$ of the simple pendulum, we can use the formula:

$$T = 2\pi\sqrt{\frac{L}{g}}$$

where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Given that $L = 1.5$ m, we can substitute this value into the formula:

$$T = 2\pi\sqrt{\frac{1.5}{9.8}}$$

Calculating this expression gives:

$$T = 2\pi\sqrt{0.153}= 0.883\, \text{s}$$

Therefore, the period $T$ of the simple pendulum is $0.883$ s.

(b) To calculate the maximum speed $v_{\text{max}}$ of the mass, we can use the conservation of mechanical energy. When the mass is at its maximum height, all gravitational potential energy is converted into kinetic energy.

Given that the mass is released from rest at an angle $\theta = 30^\circ$, its initial velocity is zero. Therefore, all the initial energy is in the form of gravitational potential energy.

The gravitational potential energy at the maximum height can be calculated using the formula:

$$PE = mgh$$

where $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the height above the equilibrium position.

Since the mass is released from rest, the height $h$ is equal to $L(1-\cos\theta)$. Substituting the known values into the equations:

$$PE = (0.2\, \text{kg})(9.8\, \text{m/s}^2)(1.5\, \text{m})(1-\cos 30^\circ)$$

Simplifying this expression gives:

$$PE = (0.2\, \text{kg})(9.8\, \text{m/s}^2)(1.5\, \text{m})(1-0.866)$$

$$PE = 1.51\, \text{J}$$

Since all the initial energy is potential energy, at the maximum speed $v_{\text{max}}$, all of it is converted into kinetic energy:

$$KE = 1.51\, \text{J}$$

We know that kinetic energy is given by the formula:

$$KE = \frac{1}{2}mv^2$$

where $m$ is the mass and $v$ is the velocity.

Solving for $v$, we get:

$$v = \sqrt{\frac{2KE}{m}}$$

Substituting the known values into this equation:

$$v_{\text{max}} = \sqrt{\frac{2(1.51\, \text{J})}{0.2\, \text{kg}}}$$

Calculating this expression gives:

$$v_{\text{max}} = \sqrt{\frac{3.02}{0.2}} = \sqrt{15.1} = 3.88\, \text{m/s}$$

Therefore, the maximum speed $v_{\text{max}}$ of the mass is $3.88$ m/s.

(c) Using energy considerations, we can calculate the maximum height $h$ above the equilibrium position that the mass reaches during its swing. This maximum height is equal to the potential energy the mass has at its maximum displacement.

Using the gravitational potential energy formula:

$$PE = mgh$$

we can solve for $h$:

$$h = \frac{PE}{mg}$$

Substituting the known values into this equation:

$$h = \frac{1.51\, \text{J}}{(0.2\, \text{kg})(9.8\, \text{m/s}^2)} = \frac{1.51\, \text{J}}{1.96\, \text{N}}$$

Calculating this expression gives:

$$h = 0.771\, \text{m}$$

Therefore, the height $h$ above the equilibrium position that the mass reaches during its swing is $0.771$ m.

(d) To calculate the tension $T_{\text{max}}$ in the string when the mass is at its maximum height, we can use the equilibrium condition. At the maximum height, the tension in the string must balance the weight of the mass.

Therefore, using the equation:

$$T_{\text{max}} - mg = 0$$

we can solve for $T_{\text{max}}$:

$$T_{\text{max}} = mg$$

Substituting the known values:

$$T_{\text{max}} = (0.2\, \text{kg})(9.8\, \text{m/s}^2)$$

Calculating this expression gives:

$$T_{\text{max}} = 1.96\, \text{N}$$

Therefore, the tension $T_{\text{max}}$ in the string when the mass is at its maximum height is $1.96$ N.

Note: Some calculations have been rounded to two decimal places for simplicity.