A weight attached to a long spring is bouncing up and down. As it bounces, its distance from the floor varies periodically with time. You start a stopwatch. When the stopwatch reads 0.3 seconds, the weight reaches its first high point 65 cm above the ground. The next low point, 25 cm above the ground, occurs at 1.8 seconds. What are the next two times the spring is at a high of 50 cm after 2.0 seconds?

Why did you raise the curve to 45 cm

let's fit the data to a curve of the type

f(t) = a cos k(t +d) + c

from highest to lowest = 65-25= 40 cm
so a = 20
from highest to lowest = .3 s to 1.8 s
so period = 3 seconds
2π/k = 3
k = 2π/3

so far we have
f(t) = 20 cos (2π/3) (t +d) + c
let's raise the curve 45 cm
f(t) = 20 cos (2π/3) (t +d) + 45

and finally , our max now occurs at t = 0, but we want to move our curve .3 to the right, so the max occurs at t=.3

f(t) = 20 cos (2π/3) (t - .3) + 45

so when does f(t) = 50 ?

20 cos (2π/3)(t - .3) + 45 = 50
cos (2π/3)(t - .3) = .25
so (2π/3)(t-.3) = 1.318116
t - .3 = .62935..
t = .92935...
or (2π/3)(t-.3) = 2π - 1.318116
t= 2.6706....

of course since the period is 3 seconds, adding or subtracting 3 seconds from any solution will yield a new solution
so how about .92935 + 3 = 3.92935 seconds ??

So the next two times after 2 seconds to reach a height of 50 cm is
t = 2.6706 and t = 3.92935

check:
f(2.6706...) = 50
f(3.92935...) = 50 , used my calculator.

To find the next two times the spring is at a high of 50 cm after 2.0 seconds, we need to determine the period of oscillation and then find the times when it reaches a height of 50 cm.

To calculate the period of oscillation, we can use the information given about the first high point and the next low point.

The period, T, can be calculated using the formula:

T = t2 - t1

Where t1 is the time when the weight reaches the first high point (0.3 seconds) and t2 is the time when the weight reaches the next low point (1.8 seconds).

T = 1.8 - 0.3
T = 1.5 seconds

Now that we know the period, we can find the frequency, f, using the formula:

f = 1 / T

f = 1 / 1.5
f = 2/3 Hz

To find the next two times the spring is at a high of 50 cm after 2.0 seconds, we need to determine the phase angle, θ, at 2.0 seconds.

The phase angle, θ, at a given time, t, can be calculated using the formula:

θ = 2πft

Where f is the frequency (2/3 Hz) and t is the given time.

θ = 2π * (2/3) * 2.0
θ ≈ 8.38 radians

Now, we can find the times when the spring is at a height of 50 cm by using the phase difference between the heights of 50 cm and 65 cm.

The phase difference, Δθ, between two heights can be calculated using the formula:

Δθ = 2π * (h1 - h2) / A

Where h1 and h2 are the two heights (65 cm and 50 cm) and A is the amplitude (half the difference between the highest and lowest points, which is (65-25)/2 = 20 cm).

Δθ = 2π * (65 - 50) / 20
Δθ ≈ 3.77 radians

To find the next two times, we need to determine the phase angle, θNext, when the height is 50 cm. We can calculate θNext using the formula:

θNext = θ + Δθ

θNext ≈ 8.38 + 3.77
θNext ≈ 12.15 radians

Now, we can find the times using the phase angle, θNext.

The formula to calculate the time, tNext, corresponding to a given phase angle is:

tNext = θNext / (2πf)

tNext ≈ 12.15 / (2π * (2/3))
tNext ≈ 5.79 seconds

So, the next two times the spring is at a high of 50 cm after 2.0 seconds are approximately 5.79 seconds.

To find the next two times the spring is at a high of 50 cm after 2.0 seconds, we need to understand the motion of the weight attached to the spring.

This type of motion is called simple harmonic motion, which follows a sinusoidal pattern. The equation for simple harmonic motion is:

y(t) = A * cos(ωt + φ) + y₀

Where:
- y(t) is the height of the weight from the ground at time t.
- A is the amplitude, which is half the difference between the maximum and minimum height.
- ω is the angular frequency, which determines the speed of oscillation.
- φ is the phase constant, which affects the starting position of the motion.
- y₀ is the vertical shift, which represents the equilibrium position of the weight (usually 0 for simplicity).

To find the values for A, ω, φ, and y₀, we can use the given information.

1. From the problem, we know that:
- At t = 0.3 seconds, the weight is at its first high point (65 cm above the ground).
- At t = 1.8 seconds, the weight is at its first low point (25 cm above the ground).

Using these two points, we can calculate the values for A, ω, φ, and y₀.

First, let's find A:
The amplitude (A) is half the difference between the maximum and minimum height.
A = (65 cm - 25 cm) / 2 = 40 cm.

Next, let's find ω:
The angular frequency (ω) can be determined using the time period (T). The time period is the time taken for one complete oscillation. In this case, we can find T by subtracting the time at the first low point from the time at the first high point.
T = 1.8 seconds - 0.3 seconds = 1.5 seconds.

Then, we can calculate ω using the formula ω = 2π / T:
ω = 2π / 1.5 ≈ 4.18879 rad/s.

Now, let's find φ:
To find the phase constant (φ), we need to consider the starting position of the motion. Since the weight is at its first high point (65 cm above the ground) at t = 0.3 seconds, we can substitute these values into the equation:
65 cm = 40 cm * cos(ω * 0.3 + φ) + y₀.

Simplifying the equation, we get:
25 cm = 40 cm * cos(ω * 0.3 + φ).

Now, solve for φ using trigonometry:
cos(ω * 0.3 + φ) = 25 cm / 40 cm.
ω * 0.3 + φ = arccos(25 / 40).
φ = arccos(25 / 40) - ω * 0.3 ≈ -1.048 rad.

Finally, let's find y₀:
As mentioned earlier, y₀ represents the equilibrium position of the weight. In this case, the weight is at its equilibrium position at t = 0 seconds, so y₀ = 0.

Now that we have the values for A, ω, φ, and y₀, we can find the next two times the spring is at a high of 50 cm after 2.0 seconds.

1. Substitute the values back into the simple harmonic motion equation:
50 cm = 40 cm * cos(ω * t + φ) + 0.

2. Solve for t:
cos(ω * t + φ) = 50 cm / 40 cm.
ω * t + φ = arccos(50 / 40).
t = (arccos(50 / 40) - φ) / ω.

Calculating t:
t₁ = (arccos(50 / 40) - (-1.048)) / 4.18879 ≈ 2.742 seconds.
t₂ = (arccos(50 / 40) - (-1.048) + 2 * π) / 4.18879 ≈ 6.326 seconds.

Therefore, the next two times the spring is at a high of 50 cm after 2.0 seconds are approximately 2.742 seconds and 6.326 seconds.