The equation of simple harmonic motion is given by:

$d(t) = 5\cos\left(6t + \frac{\pi}{6}\right)$

where:

• $d(t)$ is the displacement in meters.
• $t$ is the time in seconds.
• The amplitude is $5$ meters.
• The angular frequency is $6 \, \text{rad/s}$.

We will answer each part in turn.

---

a. Finding the earliest time when the object is farthest to the right:

The object is farthest to the right when $d(t)$ is at its maximum positive value, i.e., $d(t) = 5 \, \text{m}$.

This happens when the cosine term reaches its maximum value, which is $\cos(\theta) = 1$. So, we want:

$\cos\left(6t + \frac{\pi}{6}\right) = 1$

The cosine function is equal to 1 when the angle $\theta = 2n\pi$, where $n$ is an integer. So, set the argument of the cosine equal to $2n\pi$:

$6t + \frac{\pi}{6} = 2n\pi$

To find the earliest positive time, use $n = 0$:

$6t + \frac{\pi}{6} = 0$

Solve for $t$:

$6t = -\frac{\pi}{6}$

$t = -\frac{\pi}{36}$

Since time cannot be negative, we'll move to the next solution by setting $n = 1$:

$6t + \frac{\pi}{6} = 2\pi$

Solve for $t$:

$6t = 2\pi - \frac{\pi}{6}$

$6t = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}$

$t = \frac{11\pi}{36}$

Thus, the earliest time $t$ when the object is farthest to the right is:

$t = \frac{11\pi}{36} \, \text{seconds}$

---

b. Finding the frequency of the motion:

The angular frequency $\omega$ is given by:

$\omega = 6 \, \text{rad/s}$

The relationship between angular frequency and regular frequency $f$ is:

$\omega = 2\pi f$

Solve for $f$:

$f = \frac{\omega}{2\pi} = \frac{6}{2\pi} = \frac{3}{\pi} \, \text{Hz}$

Thus, the frequency of the motion is:

$f = \frac{3}{\pi} \, \text{Hz} \approx 0.955 \, \text{Hz}$

---

c. Finding the maximum velocity of the object:

The maximum velocity occurs when the object is at the equilibrium position, which is when the cosine term goes through 0 and the sine term reaches its maximum.

The velocity $v(t)$ is the derivative of $d(t)$:

$v(t) = \frac{d}{dt}\left(5\cos\left(6t + \frac{\pi}{6}\right)\right)$

Using the chain rule:

$v(t) = -5 \cdot 6 \sin\left(6t + \frac{\pi}{6}\right)$

$v(t) = -30 \sin\left(6t + \frac{\pi}{6}\right)$

The maximum value of $\sin(\theta)$ is 1, so the maximum velocity is:

$v_{\text{max}} = 30 \, \text{m/s}$

---

Summary of Results:

a. The earliest time at which the object is farthest to the right is $t = \frac{11\pi}{36} \, \text{seconds}$.
b. The frequency of the motion is $\frac{3}{\pi} \approx 0.955 \, \text{Hz}$.
c. The maximum velocity of the object is $30 \, \text{m/s}$.

---

Would you like further details on any step or part of this solution?

Here are 5 related questions you can explore:

1. How would you find the period of the motion?
2. How does the amplitude affect the energy in simple harmonic motion?
3. What is the acceleration of the object at the equilibrium position?
4. How would the motion change if the angular frequency doubled?
5. Can you determine the phase shift of the motion?

Tip: The maximum velocity in simple harmonic motion is always $\omega A$, where $\omega$ is the angular frequency and $A$ is the amplitude.