Chapter 7: Problem 4

Find the amplitude, period, frequency, and velocity amplitude for the motion of a particle whose distance $s$ from the origin is the given function. $$s=5 \sin (t-\pi)$$

Short Answer

Expert verified

Amplitude: 5, Period: 2π, Frequency: 1/(2π), Velocity amplitude: 5.

Step by step solution

Identify the amplitude

The amplitude of a sinusoidal function given by the form $s = A \sin(ωt + φ)$ is the coefficient in front of the sine function. Here, it's 5. So, the amplitude is 5.

Determine the angular frequency (ω)

In the given function $s=5 \sin(t-\pi)$, the term inside the sine function (t - π) indicates that the angular frequency $ω$ is 1 because it's in the form $ \sin(1 \cdot t - π)$. Therefore, $ω = 1$.

Calculate the period

The period $T$ is given by the formula $T = \frac{2\pi}{ω}$. Substituting $ω = 1$, we get $T = \frac{2\pi}{1} = 2\pi$. So, the period is $2\pi$.

Determine the frequency

The frequency $f$ is the reciprocal of the period, so $f = \frac{1}{T}$. Using the period from Step 3, $f = \frac{1}{2\pi}$.

Calculate the velocity amplitude

The velocity amplitude $V_{max}$ is given by the product of the amplitude $A$ and the angular frequency $ω$. Given $A=5$ and $ω=1$, $V_{max} = 5 \cdot 1 = 5$.

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Key Concepts

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

Amplitude

The amplitude of a sinusoidal motion describes the maximum distance that the particle moves from its origin. It's a measure of how 'tall' the waves are. In our given function, the amplitude is the coefficient in front of the sine function. For the function $$s=5 \sin(t-\pi)$$, the coefficient is 5. Therefore, the amplitude is 5 units. This means the particle moves 5 units away from the origin at its peak.

Angular Frequency

Angular frequency, denoted as $ω$, tells us how fast the particle oscillates back and forth. It's linked to the speed of the cycle. For the given function $$s=5 \sin(t-\pi)$$, since it’s in the form of $$s=A \sin(ωt + \phi)$$, we identify that the term inside the sine function affects the angular frequency. Here,$ t-\pi$ indicates that $ω = 1$ (since there is no coefficient other than 1 in front of $t$). Frequently used in physics, this unit is expressed in radians per second (rad/s).

Period

The period $T$ of the sinusoidal function is the time it takes for one full cycle of motion. To find the period, we use the formula $T = \frac{2\pi}{ω}$. With an angular frequency $ω=1$, the period is $[^2\pi]\frac{2\pi}{1} 2\pi = 2\pi$. This means the motion repeats every $2\pi$ seconds. In practical terms, after $2\pi$ seconds, the particle returns to the same position and repeats its motion.

Frequency

Frequency $f$ refers to how many cycles occur per unit time. It is the inverse of the period and tells us how often the motion repeats itself. To find the frequency, we use the formula $f= \frac{1}{T}$. With our period $2\pi$ found earlier, the frequency is $[\calbreak1}{2\pi]^$ which simplifies to \cdot5cm\$. Hence, in one second, the particle completes $\frac{1}{2\pi}$ cycles. The unit for frequency is Hertz (Hz), so we see that it oscillates approximately 0.159 cycles per second.

Velocity Amplitude

Velocity amplitude $V_{max}$ is the maximum speed the particle achieves during its motion. It's derived by multiplying the amplitude (how high the wave goes) by the angular frequency (how fast it oscillates). In our case, with an amplitude $A = 5$ and angular frequency $ω=1$, we get $V_{max} = A \cdot ω = 5 \cdot 1 = 5$. Thus, the velocity amplitude is 5 units per second, meaning at its fastest point, the particle moves 5 units/second.

Find the amplitude, period, frequency, and velocity amplitude for the motion of a particle whose distance \(s\) from the origin is the given function. $$s=5 \sin (t-\pi)$$