Question

A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small, heavy mass. How long should the rod be to make the period of the oscillations \(1.00 \mathrm{~s} ?\) a) \(0.0150 \mathrm{~m}\) b) \(0.145 \mathrm{~m}\) c) \(0.248 \mathrm{~m}\) d) \(0.439 \mathrm{~m}\) e) \(0.750 \mathrm{~m}\)

Short answer

Answer: (c) 0.248 m

Step-by-step solution

Identify the given variables and the formula

The period of the oscillations is given as \(T = 1.00 \mathrm{~s}\), and the acceleration due to gravity is \(g \approx 9.81 \mathrm{~m/s^{2}}\). We will use the formula for the period of a simple pendulum, \(T = 2\pi\sqrt{\frac{l}{g}}\).

Rearrange the formula for the length of the rod

We need to solve for \(l\). First, we need to square both sides of the equation: \(T^2 = (2\pi\sqrt{\frac{l}{g}})^2\), resulting in \(T^2 = 4\pi^2\frac{l}{g}\). Now we can rearrange the equation to get \(l = \frac{gT^2}{4\pi^2}\).

Plug in the given values and calculate the length of the rod

Now we can plug in our values and solve for \(l\). We know \(T = 1.00 \mathrm{~s}\) and \(g \approx 9.81 \mathrm{~m/s^2}\). Plugging these into the formula, we get \(l = \frac{(9.81 \mathrm{~m/s^2})(1.00 \mathrm{~s})^2}{4\pi^2}\).

Calculate the result

We can simplify the equation to get the length of the rod: \(l = \frac{9.81 \mathrm{~m}}{4\pi^2} \approx 0.248 \mathrm{~m}\).

Choose the correct answer from the given choices

Looking at the choices given, we can see that the closest answer to our calculated length of the rod is \(\boxed{\text{(c)} ~0.248 \mathrm{~m}}\).

Key concepts

Pendulum Oscillation

Understanding pendulum oscillation begins with a simple image: a weight swinging back and forth from a fixed point. This oscillation is a classic example of periodic motion, where an object moves back and forth within a certain path, repeating its movement over time.

Pendulums exhibit this type of movement in a regular, rhythmic pattern, influenced largely by the length of the rod to which the mass is attached and the force of gravity pulling the mass toward the Earth. The time it takes for one complete cycle (from one side, to the other, and back again) is known as the period of the pendulum. In a standard gravitational field, the longer the rod, the slower and wider the pendulum swings, and thus, the longer the period of oscillation.

This phenomenon is captured precisely in the formula for the pendulum period, \( T = 2\pi\sqrt{\frac{l}{g}} \). To assist students in grasping this relationship, it's useful to visualize or perform an actual pendulum experiment to see how increasing the string length affects the swinging period.

Harmonic Motion

Harmonic motion refers to the movement of an object that is periodic, and pendulum oscillation is a form of it known as simple harmonic motion (SHM), particularly when the angles involved are small. One critical aspect of SHM is that it is consistent; that is, the period remains constant regardless of the amplitude provided the oscillation remains within the small-angle approximation.

Additionally, SHM is a model for many other systems in nature and technology, such as the vibrations of a guitar string or the oscillations in electrical circuits. A deep understanding of simple harmonic motion allows students to appreciate its applications in designing clocks, musical instruments, and even in understanding the movement of playground swings.

To endow students with a practical feel for SHM, it's beneficial to showcase both physical systems and graphical representations of the motion, such as sine or cosine waves. These graphs display how an object's position changes over time in a smooth, continuous manner, analogous to the pendulum's swing.

Physics Problem Solving

Solving physics problems, like determining the required length of a pendulum rod to achieve a certain period, necessitates a structured approach. First, identify the known variables and the target variable. In this case, we know the period (\( T \)) and the acceleration due to gravity (\( g \)), while searching for the rod's length (\( l \)).

Next comes the formulation of the problem through equations that connect these variables. Once the relevant equation is selected or derived, we manipulate it algebraically to solve for the unknown variable, which in this context means isolating \( l \).

The subsequent steps involve plugging in the known values, performing the calculation, and then confirming that the result makes sense both mathematically and in the real-world context. The final step of checking the answer against multiple-choice options reinforces the importance of precision and attention to detail in physics problem-solving.

Students can enhance their problem-solving skills by working through similar problems, varying the knowns and unknowns, and understanding how changes in one variable affect the others. This kind of practice can be invaluable for developing a systematic and confident approach to tackling physics problems.