Question

A certain long pendulum, released from a height \(y_{0}\) above its rest position, will be at a height $$y=y_{0} e^{-0.75 t}$$ at \(t\) seconds. If the pendulum is released at a height of \(15 \mathrm{cm},\) at what time will the height be \(5.0 \mathrm{cm} ?\)

Short answer

The height of the pendulum will be 5.0 cm at approximately 1.465 seconds.

Step-by-step solution

Identify the given parameters

The initial height of the pendulum is given by \(y_0 = 15\) cm. We are asked to find the time \(t\) when the height is reduced to \(y = 5.0\) cm. The exponential decay formula provided is \(y=y_{0} e^{-0.75 t}\).

Set up the equation with the given values

Substitute \(y_0 = 15\) cm and \(y = 5.0\) cm into the provided equation to get \(5.0 = 15 e^{-0.75 t}\).

Isolate the exponential term

Divide both sides of the equation by 15 to isolate the exponential term, resulting in \(\frac{5.0}{15} = e^{-0.75 t}\) or \(\frac{1}{3} = e^{-0.75 t}\).

Take the natural logarithm of both sides

Apply the natural logarithm (ln) to both sides to solve for the exponent. Since \(\ln(e^x) = x\), the equation simplifies to \(\ln(\frac{1}{3}) = -0.75 t\).

Solve for the time \(t\)

Divide both sides by -0.75 to find the value of \(t\): \(t = \frac{\ln(\frac{1}{3})}{-0.75}\).

Calculate the time \(t\)

Perform the calculation using a calculator to find the value of \(t\): \(t \approx \frac{-1.0986}{-0.75}\) which is approximately \(t \approx 1.465\) seconds.

Key concepts

Exponential Decay Formula

Understanding the exponential decay formula is essential in physics, as it describes how quantities decrease over time at a rate proportional to their current value. For instance, in the given problem, the height of a pendulum at a specific time is represented by the formula
$$y=y_0 e^{-kt}$$
where:

• \(y\) is the height of the pendulum at time \(t\)
• \(y_0\) is the initial height
• \(k\) is the decay constant
• \(t\) is the time in seconds

To make the concept more accessible, think of \(k\) as a measure of how quickly the pendulum drops: a higher \(k\) means a quicker drop. It's crucial to recognize that this formula applies to various phenomena, such as radioactive decay and cooling processes, making it a versatile tool in physical sciences.

Natural Logarithm Applications

The natural logarithm, often denoted as 'ln', is a powerful mathematical tool with widespread applications in physics, especially when dealing with exponential decay. The natural logarithm is the inverse of the exponential function, meaning that for any positive number \(x\), \(\textrm{ln} (e^x) = x\). This property is extremely useful when you need to solve an equation for the exponent, as seen in the given exercise.
In step 4 of the solution, taking the natural logarithm of both sides transforms the equation from its exponential form to a linear one, making it possible to isolate the variable \(t\). This maneuver is a common technique used not just in academic exercises, but also in real-world scenarios such as calculating the half-life of a radioactive substance or determining the time needed for a chemical reaction to reach a certain completion level.

Solving Exponential Equations

Solving exponential equations is a fundamental skill in physics, as it allows us to determine key variables in processes that involve growth or decay. The steps to solve an exponential equation typically involve isolating the exponential term and then applying the natural logarithm to both sides of the equation. Here's a simplified guide on how to approach these problems:

• Isolate the exponential term: Get the term with the exponent by itself on one side of the equation.
• Apply the natural logarithm: Take the natural logarithm of both sides. Remember, \(\textrm{ln}(e^x) = x\).
• Solve for the unknown: With the exponent now by itself, you can solve for the variable of interest.

When applying these steps to the original exercise, the equation is rearranged to isolate \(t\), the time at which the pendulum reaches a certain height. By mastering this process, students can tackle a wide range of problems, from simple textbook questions to complex real-life scenarios involving exponential relationships.