91Ó°ÊÓ

In wave motion, the transverse speed represents the rate at which a point on a wave moves up and down, horizontally across a medium. This movement is perpendicular to the direction of wave propagation. To find the transverse speed of a wave described by the function \(y(x, t)=(15.0 \, \text{cm}) \cos (\pi x-15 \pi t)\), we need to take the derivative of this wave function with respect to time \(t\). The formula is simple to understand. Derivating the function with respect to \(t\) gives us the transverse speed:

• \(\frac{\partial y}{\partial t} = -15 \pi \times 15.0 \, \text{cm} \times \sin(\pi x - 15\pi t)\),

which indicates how fast the point moves across time. The speed will be negative or positive depending on the direction of the movement.By understanding this concept, you can grasp how the speed is determined by oscillation within the wave, calculated using the partial derivative of the wave function.

The wave equation is a cornerstone in physics, providing a mathematical description of how wave phenomena manifest in different mediums. In this context, the wave equation yields a pattern to describe a vibrating string. The general form of the wave equation for a stretched string is given by:

• \(y(x, t) = A \cos(kx - \omega t)\)

where \(A\) stands for amplitude, \(k\) for wave number, and \(\omega\) for angular frequency.In the given problem, \(y(x, t) = (15.0 \, \text{cm}) \cos(\pi x - 15\pi t)\) represents a traveling wave where the maximum amplitude is 15.0 cm. The terms \(\pi x\) and \(15\pi t\) play crucial roles:

• \(\pi\) signifies the wave number \(k\). It determines how many complete wave cycles fit into a given length.
• \(15\pi\) links to \(\omega\), describing how quickly the wave oscillates over time.

The wave equation can be solved to find quantities such as maximum displacement, speed, or period of the wave.

Partial derivatives are an essential tool in calculus, especially when dealing with functions of multiple variables. In the context of wave motion, they help us explore how wave functions change over space \(x\) and time \(t\).For the wave function \(y(x, t) = (15.0 \, \text{cm}) \cos(\pi x - 15\pi t)\), we calculate the following partial derivatives:

• \(\frac{\partial y}{\partial x}\): This indicates how the wave displacement \(y\) changes with respect to position \(x\).
• \(\frac{\partial y}{\partial t}\): This tells us how the wave displacement \(y\) changes over time, which is related to the transverse speed.

In this exercise, focusing on the partial derivative with respect to time provides insight into the wave's movement across the medium (transverse speed). Remember, a partial derivative hones in on how one variable changes while holding others constant, allowing a closer look at the dynamics involved.

Wave function analysis involves dissecting the wave equation to identify crucial characteristics like amplitude, frequency, period, wave speed, and more. For the function \(y(x, t)=(15.0 \, \text{cm}) \cos(\pi x-15 \pi t)\), let's break down each component:

• Amplitude (A): The amplitude of the wave, here 15.0 cm, shows the maximum displacement from the equilibrium position.
• Wave Number (k): Given by \(\pi\), it signifies the spatial frequency of the wave, indicating how many wavelengths fit into a unit distance.
• Angular Frequency (\(\omega\)): Represented by \(15\pi\), it defines how fast the wave oscillates over time.
• Phase (\(\theta\)): The term inside the cosine, \(\pi x - 15\pi t\), describes the phase of the wave, crucial in determining the wave's precise value at any point \((x, t)\).

Analyzing such a function is vital in understanding the behavior of waves, noting how changes in variables affect the wave's shape, speed, and direction.