a particle moves in a straight line such that its position x from a fixed point 0 at time 't' is given by

x= 5 + 8sin2t + 6cos2t

1. Find the period and amplitude of the particle.

2. Find the greatest speed of the particle.

Thanks

amplitude=10

period is also 2pi

dx/dt = 16cos2t - 12sin2t
amplitude = 20, so that would be its maximum speed.

Can you please explain how to get the answers please?

To find the period and amplitude of the particle's motion, we'll need to rewrite the equation in the form:

x = A + Bsin(ωt + φ)

where A is the amplitude, B is a coefficient, ω is the angular frequency, and φ is the phase angle.

Comparing the given equation with the standard form, we have:

x = 5 + 8sin(2t) + 6cos(2t) = A + Bsin(ωt + φ)

From the equation, we can see that:

A = 5, B = √(8^2 + 6^2) = √100 = 10

The amplitude is the magnitude of the coefficient of the sine and cosine terms.

Now, let's express the equation in terms of sine and cosine components:

x = 10(cos(φ)sin(ωt) + sin(φ)cos(ωt))

Comparing this with the standard form, we get:

A = 10cos(φ) and B = 10sin(φ)

Using these relationships, we can solve for the phase angle φ:

tan(φ) = B/A

tan(φ) = (10sin(φ))/(10cos(φ))

tan(φ) = sin(φ)/cos(φ)

tan(φ) = 1

The tangent of the phase angle φ is equal to 1. This implies that φ = π/4 or 45 degrees in the first quadrant.

Now, we know that the angular frequency is related to the period T by the formula:

ω = 2π/T

From the given equation, we see that ω = 2.

Substituting the values of A, B, and φ into the equation, we get:

x = 10cos(π/4)sin(2t + π/4)

Therefore, the period T is given by:

T = 2π/ω = 2π/2 = π

Hence, the period is equal to π.

To find the greatest speed of the particle, we can compute the derivative of the equation of motion:

v = dx/dt = d/dt(5 + 8sin(2t) + 6cos(2t))

v = 16cos(2t) - 12sin(2t)

To find the maximum value of v, we differentiate v with respect to t again:

a = dv/dt = d/dt(16cos(2t) - 12sin(2t))

a = -32sin(2t) - 24cos(2t)

Setting a = 0, we find the critical points:

-32sin(2t) - 24cos(2t) = 0

Dividing by cos(2t), we get:

-32tan(2t) - 24 = 0

tan(2t) = -24/32 = -3/4

Taking the arctangent of both sides, we find:

2t = arctan(-3/4)

t = (1/2) * arctan(-3/4)

Note: The arctangent function has multiple solutions, but we only need one solution within the interval for the maximum speed.

To find the greatest speed, we evaluate v at this particular t:

v = 16cos(2t) - 12sin(2t)

v = 16cos(arctan(-3/4)) - 12sin(arctan(-3/4))

v ≈ -19.47

Therefore, the greatest speed of the particle is approximately 19.47 units per unit time.

To find the period and amplitude of the particle's motion, we need to examine the given equation:

x = 5 + 8sin(2t) + 6cos(2t)

1. Finding the period:
The period of a trigonometric function can be determined using the equation 2π/ω, where ω is the coefficient of t in the sine or cosine function. In this case, we have sin(2t) and cos(2t), so the coefficient is 2. Therefore, the period T can be calculated as T = 2π/ω = 2π/2 = π. The period of the particle's motion is π.

2. Finding the amplitude:
The amplitude of a trigonometric function can be found by calculating the square root of the squares of the coefficients of the sine and cosine functions. In this case, the coefficient of sin(2t) is 8, and the coefficient of cos(2t) is 6. So, the amplitude A can be computed as A = √(8² + 6²) = √(100) = 10. The amplitude of the particle's motion is 10.

Now, let's move on to finding the greatest speed of the particle.

To find the greatest speed, we need to determine the magnitude of the velocity. The velocity v can be obtained by taking the derivative of the position x with respect to time t:

v = dx/dt = d(5 + 8sin(2t) + 6cos(2t))/dt.

Differentiating each term using the chain rule, we obtain:

v = 2(8cos(2t) - 6sin(2t)).

Now, to find the greatest speed, we need to find the maximum value of v. To do that, we find when the derivative of v with respect to t is equal to zero:

dv/dt = d(2(8cos(2t) - 6sin(2t)))/dt = -32sin(2t) - 24cos(2t) = 0.

Simplifying the equation, we get:

-8sin(2t) - 6cos(2t) = 0.

We can divide the equation by -2 to simplify it further:

4sin(2t) + 3cos(2t) = 0.

To solve this equation, we can rewrite it in the form of tan(θ):

tan(θ) = -4/3.

Using the inverse tangent function, we find:

θ = arctan(-4/3) ≈ -0.93 radians.

Since we are looking for the greatest speed, we consider the case when sin(2t) is at its maximum value (1). Therefore, we substitute sin(2t) = 1 back into the equation for v:

v = 2(8cos(2t) - 6sin(2t)) = 2(8cos(2t) - 6) = 16cos(2t) - 12.

We can evaluate this when cos(2t) is at its maximum value (1):

v = 16(1) - 12 = 4.

Hence, the greatest speed of the particle is 4 units per time.