Question

Find the amplitude, period, frequency, and velocity amplitude for the motion of a particle whose distance \(s\) from the origin is the given function. $$ s=2 \sin (4 t-1) $$

Short answer

Amplitude = 2, Period = \( \frac{\pi}{2} \), Frequency = \( \frac{2}{\pi} \), Velocity amplitude = 8

Step-by-step solution

Identify the general form

The general form for simple harmonic motion is given by the equation \[ s = A \sin( \omega t + \phi ) \] where - \( A \) is the amplitude, - \( \omega \) is the angular frequency, - \( t \) is time, and - \( \phi \) is the phase shift.

Compare the given function

Compare the given function \[ s = 2 \sin(4t - 1) \] with the general form. Here, you can identify: - \( A = 2 \) - \( \omega = 4 \) - \( \phi = -1 \)

Find the amplitude

The amplitude \( A \) is the coefficient of the sine function in the given equation. So, the amplitude is \( 2 \).

Calculate the period

The period \( T \) of the motion is given by the formula \[ T = \frac{2\pi}{\omega} \] Substituting \( \omega = 4 \), you get \[ T = \frac{2\pi}{4} = \frac{\pi}{2} \]

Determine the frequency

The frequency \( f \) is the reciprocal of the period \[ f = \frac{1}{T} \] Substituting \( T = \frac{\pi}{2} \), you get \[ f = \frac{2}{\pi} \]

Find the velocity amplitude

The velocity amplitude \( V_A \) is found by multiplying the amplitude \( A \) by the angular frequency \( \omega \). So, \[ V_A = A \cdot \omega = 2 \cdot 4 = 8 \]

Key concepts

Amplitude Calculation

In the context of simple harmonic motion, the amplitude represents the maximum displacement of the particle from its equilibrium position. For the given function
s = 2 \sin(4t - 1), we determine the amplitude by identifying the coefficient of the sine function. Here, the coefficient is 2.

• The amplitude (\(A\)) of the particle's motion is 2.

This value can be interpreted as the furthest distance the particle moves away from the origin in its oscillatory motion.

Angular Frequency

Angular frequency (\(\omega\)) is a measure of how quickly the particle oscillates through its cycle. It can be extracted directly from the argument of the sine function in our equation. The standard form for simple harmonic motion is
s = A \sin(\omega t + \phi), and comparing it with
s = 2 \sin(4t - 1), we recognize that
\(\omega = 4\).

• Angular Frequency (\(\omega\)) = 4 radians per second

The higher the angular frequency, the faster the particle completes one oscillation.

Period Determination

The period (\(T\)) of harmonic motion is the time it takes for the particle to complete one full cycle of its motion. We can calculate the period using the formula: \[T = \frac{2\pi}{\omega} \]. Substituting our previously found value of \(\omega\) = 4, we get \[ T = \frac{2\pi}{4} = \frac{\pi}{2} \].

• Period (\(T\)) = \frac{\pi}{2} seconds

This means it takes approximately 1.57 seconds for the particle to return to its initial position.

Frequency Calculation

The frequency (\(f\)) is the number of oscillations the particle completes per second. It is the reciprocal of the period: \[f = \frac{1}{T}\]. By substituting the period from our previous calculation (\(T = \frac{\pi}{2} \)), we get: \[ f = \frac{2}{\pi} \].

• Frequency (\(f\)) \approx 0.637 Hz

This shows that the particle completes just over half an oscillation every second.

Velocity Amplitude

Velocity amplitude (\(V_A\)) represents the maximum speed of the particle during its motion. It can be calculated by multiplying the amplitude (\(A\)) by the angular frequency (\(\omega\)): \[ V_A = A \cdot \omega \]. Using our previously determined values of \(A = 2\) and \(\omega = 4\), we find:\[ V_A = 2 \cdot 4 = 8 \].

• Velocity Amplitude (\(V_A\)) = 8 units per second

Therefore, the particle's maximum speed is 8 units per second during its oscillations.