An object moves with simple harmonic motion of amplitude A and period T. During one complete oscillation, for what length of time is the object's distance from equilibrium greater than A/2? Enter your answer as a formula involving the period.

Question

An object moves with simple harmonic motion of amplitude A and period T. During one complete oscillation, for what length of time is the object's distance from equilibrium greater than A/2? Enter your answer as a formula involving the period.

How much time does it take the object to cover a total distance of 4A? Enter your answer as a formula involving the period.

I've got the first part of the

Answer 1

It takes one period (T) for the object to travel 4A. That's 2A going positive and back to zero, and 2A going negative and back to zero. (Starting at zero)

For the first question, assume

Y = A sin(2 pi t/T)

How much of the time is Y > A/2?

Do the math. Find the times t when y = A/2. The figure out the times when y > A/2, between those points in a single cycle.

Answer 2

To find the length of time the object's distance from equilibrium is greater than A/2, we need to determine the phase of the motion.

In simple harmonic motion, the displacement from equilibrium at any given time t can be described by the equation:

x(t) = A * cos(2πt/T + φ)

where x(t) is the displacement, A is the amplitude, T is the period, t is the time, and φ is the phase constant.

Since we want to find the time when the distance from equilibrium is greater than A/2, we can set up the following inequality:

|A * cos(2πt/T + φ)| > A/2

Now, solving this inequality for time t will give us the length of time the object's distance from equilibrium is greater than A/2.

First, let's consider the positive side of the inequality:

A * cos(2πt/T + φ) > A/2

Dividing both sides by A:

cos(2πt/T + φ) > 1/2

Now, we need to find the values of t that satisfy this inequality. The region where cos(2πt/T + φ) > 1/2 is when the angle (2πt/T + φ) lies within the first and fourth quadrants of the unit circle.

The angle in the first quadrant where cosθ = 1/2 is π/3, and in the fourth quadrant is 5π/3.

So, we need to solve the following inequality:

0 ≤ 2πt/T + φ < π/3 or 5π/3 ≤ 2πt/T + φ < 2π

Subtracting φ and rearranging:

-φ ≤ 2πt/T < π/3 - φ or 5π/3 - φ ≤ 2πt/T < 2π - φ

Now, we can solve each of these inequalities separately:

For -φ ≤ 2πt/T < π/3 - φ,

we can subtract φ and multiply by T/(2π):

-Tφ/(2π) ≤ t < (T/2π)(π/3 - φ)

Similarly, for 5π/3 - φ ≤ 2πt/T < 2π - φ, we get:

(T/2π)(5π/3 - φ) ≤ t < T - (Tφ/(2π))

Therefore, the total time the object's distance from equilibrium is greater than A/2 is:

[(T/2π)(π/3 - φ)] - [-Tφ/(2π)] + [T - (Tφ/(2π))]

Simplifying this expression, we get:

(T/2π)(π/3 + 5π/3) + [T - (-Tφ/(2π))]

(T/2π)(2π) + [T + Tφ/(2π)]

T + Tφ/(2π) + T - Tφ/(2π)

2T

So, the length of time the object's distance from equilibrium is greater than A/2 is 2T.

Now, let's move on to the second part of the question.

Answer 3

solution for you. To find the length of time the object's distance from equilibrium is greater than A/2, we can start by considering the equation of simple harmonic motion:

x(t) = A * cos(2πt/T)

Here, x(t) represents the position of the object at time t, A is the amplitude, T is the period, and cos is the cosine function.

To find the length of time the object's distance from equilibrium is greater than A/2, we need to find the values of t where |x(t)| > A/2. Since cos(θ) is positive for angles θ between -π/2 and π/2, we can rewrite the inequality as:

A/2 < A * cos(2πt/T) < A/2

Dividing both sides of the inequality by A, we get:

1/2 < cos(2πt/T) < 1/2

Now, we can take the inverse cosine of both sides:

arccos(1/2) < 2πt/T < arccos(1/2)

Simplifying, we have:

π/3 < 2πt/T < π/3

Now, we can solve for t by dividing each side of the inequality by 2π and multiplying by T:

(π/3) * (T/2π) < (2πt/T) * (T/2π) < (π/3) * (T/2π)

Simplifying further:

T/6 < t < T/6

So, the object's distance from equilibrium is greater than A/2 for a duration of T/6.

Now, let's move on to the second part of the question: how much time does it take the object to cover a total distance of 4A?

The total distance covered by the object in one complete oscillation is 2A, since it moves from one extreme point to the other. Therefore, to cover a total distance of 4A, the object will need to complete two oscillations.

Since the period T represents the time it takes for one complete oscillation, we can conclude that the time it takes for the object to cover a total distance of 4A is 2T.

Therefore, the formula involving the period to calculate the time t is:

t = 2T