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A point moves along the \(x\)-axis according to the equation \(x=a \sin ^{2}\left(\omega t-\frac{\pi}{4}\right)\). Find the amplitude, period, velocity and projection \(v_{x}\) as a function of \(x\).

Short Answer

Expert verified
Amplitude: \(a\), Period: \(\frac{2\pi}{\omega}\), Velocity: \(v_{x} = \pm a \omega \sqrt{1 - \frac{x}{a}}\).

Step by step solution

01

Identify the Function Components

The given position function is \( x = a \sin^{2}\left(\omega t - \frac{\pi}{4}\right) \). From this, we can see that the maximum value of \( \sin^{2} \) is 1, which means \( x \) can vary from 0 to \( a \) in maximum amplitude.
02

Calculate the Amplitude

The amplitude of the motion is the maximum displacement from the mean position. Since the maximum value of \( \sin^{2} \) is 1, the amplitude of \( x \) is \( a \).
03

Determine the Period

To find the period, observe that the angle \( \theta = \omega t - \frac{\pi}{4} \) completes a full cycle when it increases by \( 2\pi \). Therefore, \( 2\pi = \omega T \), yielding the period \( T = \frac{2\pi}{\omega} \).
04

Find the Velocity Expression

To find the velocity, differentiate \( x \) with respect to \( t \). Using the chain rule, \( v = \frac{dx}{dt} = 2a \sin\left(\omega t - \frac{\pi}{4}\right) \cos\left(\omega t - \frac{\pi}{4}\right) \cdot \omega \). This simplifies to \( v = a \omega \sin\left(2\left(\omega t - \frac{\pi}{4}\right)\right) \).
05

Express the Velocity in Terms of x

Since \( x = a \sin^{2}\left(\omega t - \frac{\pi}{4}\right) \), let \( \sin\left(\omega t - \frac{\pi}{4}\right) = y \). Then \( x = a y^{2} \) or \( y^{2} = \frac{x}{a} \), leading to \( y = \pm \sqrt{\frac{x}{a}} \). Substitute into the velocity expression: \( v = \pm a \omega \sqrt{1 - \frac{x}{a}} \). Therefore, the projection of velocity on the \( x \)-axis is \( v_{x} = \pm a \omega \sqrt{1 - \frac{x}{a}} \).

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Key Concepts

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

Amplitude of Motion
The amplitude of motion is a crucial concept to understand in simple harmonic motion. It represents the maximum extent of the oscillation from the mean or central position. In this particular problem, the position of the point is given as \( x = a \sin^{2}\left(\omega t - \frac{\pi}{4}\right) \). The function \( \sin^{2} \) has a maximum possible value of 1. Therefore, the maximum value that \( x \) can take, which is the "amplitude," equals \( a \). This means the point can move a maximum distance of \( a \) from the mean position on the \( x \)-axis.

The amplitude is thought of as the "height" or "strength" of the oscillation, telling us how far the motion extends in positive or negative directions from its equilibrium position. It is essential for predicting the behavior and limitations of oscillating systems.
Period of Oscillation
The period of oscillation is another fundamental aspect of simple harmonic motion. It tells us how long it takes for the system to complete one full cycle of motion. In the given motion equation, \( x = a \sin^{2}\left(\omega t - \frac{\pi}{4}\right) \), one full cycle corresponds to the angle \( \omega t - \frac{\pi}{4} \) increasing by \( 2\pi \).

Thus, we set the equation \( 2\pi = \omega T \) to find the period \( T \). Solving for \( T \), we find that \( T = \frac{2\pi}{\omega} \). This result shows that the period is inversely proportional to \( \omega \), the angular frequency. In simpler terms, if the speed of oscillation (given by \( \omega \)) increases, the period decreases, meaning the cycle completes more quickly.

Understanding the period is vital for synchronizing systems and predicting when an oscillating object will be in certain positions.
Velocity Function
The velocity function is essential for understanding how fast a particle is moving at any point in time and direction during its motion. To obtain the velocity, we need to differentiate the position function with respect to time \( t \). Given \( x = a \sin^{2}\left(\omega t - \frac{\pi}{4}\right) \), we use the chain rule for differentiation.

The derivative yields \( v = \frac{dx}{dt} = 2a \sin\left(\omega t - \frac{\pi}{4}\right) \cos\left(\omega t - \frac{\pi}{4}\right) \cdot \omega \). Using the trigonometric identity, this simplifies to \( v = a \omega \sin\left(2(\omega t - \frac{\pi}{4})\right) \).

This expression gives us the velocity at any time \( t \). It's crucial for analyzing how quickly the object is moving and can predict timing and behavior in dynamic systems.
Projection of Velocity
The concept of the projection of velocity helps us express the velocity of a particle in terms of its position \( x \). This is a key concept when exploring practical applications of simple harmonic motion. From the given equation, \( x = a \sin^{2}\left(\omega t - \frac{\pi}{4}\right) \), let us assume \( y = \sin\left(\omega t - \frac{\pi}{4}\right) \), then \( x = a y^{2} \) leads to \( y^{2} = \frac{x}{a} \)

This means \( y = \pm \sqrt{\frac{x}{a}} \). Substitute \( y \) back into the velocity equation to express it in terms of \( x \). We find that \( v = \pm a \omega \sqrt{1 - \frac{x}{a}} \).

In this final expression, the velocity \( v \) is a function of \( x \), which can help determine how the particle moves along the \( x \)-axis at any given point. This aspect of simple harmonic motion is particularly useful in understanding real-world scenarios where the position affects the object's velocity, like springs or pendulum movements.

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Most popular questions from this chapter

Temperature dependence of speed of sound could be expressed as, \(v=k \sqrt{T}\), where \(k\) is some positive constant, In a medium sound is travelling along \(x\)-axis. At \(x=0\), temperature is \(T_{1}\) and at \(x=l\), temperature is \(T_{2}\) and the temperature varies linearly with \(x\). Time taken by sound to travel from \(x=0\) to \(x=l\) is (a) \(\frac{2 l}{k\left(\sqrt{T_{1}}+\sqrt{T_{2}}\right)}\) (b) \(\frac{l}{k\left(\sqrt{T_{1}}+\sqrt{T_{2}}\right)}\) (c) \(\frac{l}{2 k\left(\sqrt{T_{1}}+\sqrt{T_{2}}\right)}\) (d) \(\frac{l}{2 k\left(\sqrt{T_{1}}+\sqrt{T_{2}}\right)^{2}}\)

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