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The displacement from equilibrium of an oscillating mass attached to a spring is given by \(y(t)=4 \cos 3 \pi t\) where \(y\) is measured in inches and \(t\) in seconds. Find the displacement at the times indicated in the table. $$\begin{array}{|c|c|}\hline t & y(t) \\\\\hline 0 & \\\0.25 & \\\0.50 & \\\0.75 & \\\1.00 & \\\1.25 & \\ \hline\end{array}$$

Short Answer

Expert verified
At \( t = 0, y = 4 \); at \( t = 0.25, y = 0 \); at \( t = 0.50, y = -4 \); at \( t = 0.75, y = 0 \); at \( t = 1.00, y = -4 \); at \( t = 1.25, y = 0 \).

Step by step solution

01

Interpret the given function

The given function for the displacement is \(y(t) = 4\cos(3\pi t)\). Here, \(y(t)\) is the displacement of the mass from the equilibrium position at time \(t\). The amplitude is 4 inches and the angular frequency is \(3\pi\).
02

Calculate displacement at t = 0

Substitute \(t = 0\) into the displacement function: \[y(0) = 4\cos(3\pi \cdot 0) = 4\cos(0) = 4\times 1 = 4\]Thus, the displacement at \(t=0\) is 4 inches.
03

Calculate displacement at t = 0.25

Substitute \(t = 0.25\) into the displacement function:\[y(0.25) = 4\cos(3\pi \times 0.25) = 4\cos(0.75\pi)\]\(\cos(0.75\pi)\) is equal to zero, so \[y(0.25) = 4\times 0 = 0\]Thus, the displacement at \(t=0.25\) is 0 inches.
04

Calculate displacement at t = 0.50

Substitute \(t = 0.50\) into the displacement function:\[y(0.50) = 4\cos(3\pi \times 0.50) = 4\cos(1.5\pi)\]\(\cos(1.5\pi)\) is equal to negative one, so \[y(0.50) = 4 \times (-1) = -4\]Thus, the displacement at \(t=0.50\) is -4 inches.
05

Calculate displacement at t = 0.75

Substitute \(t = 0.75\) into the displacement function:\[y(0.75) = 4\cos(3\pi \times 0.75) = 4\cos(2.25\pi)\]\(\cos(2.25\pi)\) is equal to zero, so \[y(0.75) = 4 \times 0 = 0\]Thus, the displacement at \(t=0.75\) is 0 inches.
06

Calculate displacement at t = 1.00

Substitute \(t = 1.00\) into the displacement function:\[y(1.00) = 4\cos(3\pi \times 1.00) = 4\cos(3\pi)\]\(\cos(3\pi)\) is equal to negative one, so \[y(1.00) = 4 \times (-1) = -4\]Thus, the displacement at \(t=1.00\) is -4 inches.
07

Calculate displacement at t = 1.25

Substitute \(t = 1.25\) into the displacement function:\[y(1.25) = 4\cos(3\pi \times 1.25) = 4\cos(3.75\pi)\]\(\cos(3.75\pi)\) is equal to zero, so \[y(1.25) = 4 \times 0 = 0\]Thus, the displacement at \(t=1.25\) is 0 inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Oscillations
In the context of physics, oscillations refer to repeated back-and-forth movements around an equilibrium position. A common example is the motion of a mass attached to a spring. This motion can be characterized by how the displacement of the mass changes over time. Oscillatory motion is observed in various scenarios, from the vibration of guitar strings to the swinging of a pendulum.
The function given, \(y(t) = 4\cos(3\pi t)\), is a perfect representation of oscillation, as it describes how the displacement of the mass from its resting position changes over time. The oscillation occurs in a sinusoidal manner, meaning the displacement follows a predictable wave-like pattern.
Exploring the Cosine Function
The cosine function is a fundamental trigonometric function used to model oscillatory behavior. In simple terms, it describes a wave that starts at its maximum value. This function is particularly useful when representing periodic phenomena, where a wave returns to its initial state after a fixed interval.
The equation \(y(t) = 4\cos(3\pi t)\) uses the cosine function to illustrate displacement over time. Here, the argument of the cosine function \(3\pi t\) determines the position on the wave at any given time \(t\). The output of the cosine function then scales according to the amplitude, representing the height of the oscillations.
  • The cosine function starts at 1 when time is zero.
  • It reaches zero when \(t\) is a quarter of its period.
  • It continues to -1 at half the period, back to zero at three-quarters, and returns to 1 at full period.
Deciphering Amplitude
Amplitude is a crucial aspect of any oscillation function. It refers to the maximum distance from the equilibrium position during oscillation. Think of it as half the distance between the highest and lowest points of the wave. In the function \(y(t) = 4\cos(3\pi t)\), the amplitude is 4 inches.
The amplitude tells us how far the mass will move away from its resting, or equilibrium, position. Larger amplitude means larger oscillations and vice versa. It is important to understand that amplitude solely determines the peak value of oscillation and does not influence the frequency or speed of oscillations.
Understanding Angular Frequency
Angular frequency is a measure of how quickly an object oscillates. It is denoted by the Greek letter omega (\(\omega\)) and is related to the regular frequency, which tells us how many oscillations occur in one second. Angular frequency considers the concept of one full cycle being \(2\pi\) radians.
In our given equation, \(y(t) = 4\cos(3\pi t)\), the angular frequency is \(3\pi\). This value determines how many radians per second the oscillation covers. A higher angular frequency means the mass attached to the spring oscillates back and forth more rapidly. This attribute directly influences the speed of oscillation but not the extent of each oscillation, which is governed by the amplitude.
  • Angular frequency \(\omega\) is given in radians per second.
  • It is connected to the period \(T\) of the motion by \(\omega = \frac{2\pi}{T}\).

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