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Chapter 9: Problem 48

The equation of motion of a particle is \(x=a \cos (\alpha t)^{2}\). The motion is (A) periodic but not oscillatory. (B) periodic and oscillatory. (C) oscillatory but not periodic. (D) neither periodic nor oscillatory.

Short Answer

Expert verified
The motion of the particle is described by the equation \(x = a \cos(\alpha t^2)\). Analyzing the argument of the cosine function, we observe that as time increases, the argument increases quadratically. This causes the time it takes to complete a cycle to decrease as time progresses. Therefore, the motion is not periodic. However, since the cosine function causes oscillations in the motion, the particle's motion is oscillatory but not periodic. The correct option is (C) oscillatory but not periodic.

Step by step solution

01

Analyze the equation of motion

To do this, let us first write down the given equation of motion. It is given by: \(x = a \cos(\alpha t^2)\)
02

Investigate the nature of the cosine function

The cosine function is a periodic function having a period of \(2 \pi\). It oscillates between -1 and 1. Therefore, any function of the form \(x = a \cos(\beta)\), where a and 尾 are constants, will be periodic and oscillatory.
03

Observe the argument of the cosine function

In the given equation, the argument of the cosine function is \(\alpha t^2\), where \(\alpha\) is a constant and t is time. Notice that the argument is the square of time which implies that as time increases, the argument will also increase quadratically.
04

Analyze the behavior of the argument

In order to see how the argument changes with time let's consider two cases: a) For \( 0 \leq \alpha t^2 \< \pi\), the value of \(\cos(\alpha t^2)\) will go through one complete cycle from 1 to -1 and back to 1. b) For \(\pi \leq \alpha t^2 \< 2 \pi\), the value of \(\cos(\alpha t^2)\) will go through two complete cycles since now the argument will move from \(\pi\) to \(2 \pi\). This pattern continues as time goes on, meaning the time it takes to complete a cycle decreases as time progresses.
05

Conclude the type of motion

From the analysis, we can see that the motion of the particle is not periodic since the time it takes to complete a cycle keeps reducing. However, the cosine function does cause oscillations in the motion. Therefore, the motion of the particle is oscillatory but not periodic.
06

Answer

So, the correct option is (C) oscillatory but not periodic.

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