For the next part, you will need to know that the position d of an object attached to a spring that oscillates with period T=1.0s is a sinusoidal function of time: d=kcos(2πt)d=kcos(2πt) for some constant k.

Assume that the initial position of the object is d0=2.0cm. Find the following positions of the object: d1 at t1=0.25s , d2 at t2=0.50s , d3 at t3=0.75s, and d4 at t4=1.0s.

so I found out k=2cm. I plug everything in and got 1.999, 1.997, 1.993, and 1.988.
masteringphysics says im still wrong?
Help please!

By looking at the sinusoidal wave graphed out, the period is T=1.0s. Which means in one second the d will be the same as the beginning. So we know that d0=2.0cm so d4 at 1.0s = d0 or 2cm. Then we know that half of that period would be the negative of the intial or d2 at .5s= -2.0cm. Finally the remaining times are half of that which will make them rest on the axis = 0. d1=0,d2=-2,d3=0,d4=2.

James' response was accurate

Well, it seems like there's a problem with your calculations. Let's double-check your math.

We know that the equation for the position of the object is d = kcos(2πt), where d is the position, k is the constant, and t is the time.

Given that k = 2cm and the initial position d0 = 2.0cm, let's find the positions at different times:

For t1 = 0.25s:
d1 = kcos(2π*0.25) = 2cos(0.5π) = 2cm

For t2 = 0.50s:
d2 = kcos(2π*0.50) = 2cos(π) = -2cm

For t3 = 0.75s:
d3 = kcos(2π*0.75) = 2cos(1.5π) = -2cm

For t4 = 1.0s:
d4 = kcos(2π*1.0) = 2cos(2π) = 2cm

So, the correct positions of the object are: d1 = 2cm, d2 = -2cm, d3 = -2cm, and d4 = 2cm.

If you got different answers, then there might have been an error in your calculations. Double-check your calculations and try again. Good luck!

To find the positions of the object at different times, you need to substitute the respective values of time into the equation d = kcos(2πt) and calculate the result.

Let's go step-by-step to find the positions at each given time:

1. For t1 = 0.25s:
Substitute t1 into the equation: d1 = kcos(2π(0.25))
Calculate: d1 = kcos(0.5π)

2. For t2 = 0.50s:
Substitute t2 into the equation: d2 = kcos(2π(0.50))
Calculate: d2 = kcos(π)

3. For t3 = 0.75s:
Substitute t3 into the equation: d3 = kcos(2π(0.75))
Calculate: d3 = kcos(1.5π)

4. For t4 = 1.0s:
Substitute t4 into the equation: d4 = kcos(2π(1.0))
Calculate: d4 = kcos(2π)

Now, let's substitute the given value k = 2cm and calculate the positions:

1. For t1 = 0.25s:
d1 = 2cos(0.5π)
d1 ≈ 2(0.7071)
d1 ≈ 1.414 cm

2. For t2 = 0.50s:
d2 = 2cos(π)
d2 ≈ 2(-1)
d2 ≈ -2 cm

3. For t3 = 0.75s:
d3 = 2cos(1.5π)
d3 ≈ 2(-0.7071)
d3 ≈ -1.414 cm

4. For t4 = 1.0s:
d4 = 2cos(2π)
d4 ≈ 2(1)
d4 ≈ 2 cm

Therefore, the correct positions of the object are approximately:
d1 = 1.414 cm, d2 = -2 cm, d3 = -1.414 cm, and d4 = 2 cm.

Please double-check your calculations and ensure that you are using the correct formula and values.

To find the positions of the object at different times, you need to use the equation d = kcos(2πt), where d is the position of the object, k is the constant, and t is the time.

First, let's find the value of k based on the given initial position d0 = 2.0 cm. Substitute d0 into the equation and solve for k:

2.0 cm = kcos(2π(0))
2.0 cm = kcos(0)
2.0 cm = k

Therefore, k = 2.0 cm.

Now, we can find the positions of the object at different times using the formula d = kcos(2πt).

For t1 = 0.25 s:
d1 = 2.0 cm * cos(2π * 0.25 s)
d1 = 2.0 cm * cos(π/2)
d1 = 2.0 cm * 0
d1 = 0 cm

For t2 = 0.50 s:
d2 = 2.0 cm * cos(2π * 0.50 s)
d2 = 2.0 cm * cos(π)
d2 = 2.0 cm * (-1)
d2 = -2.0 cm

For t3 = 0.75 s:
d3 = 2.0 cm * cos(2π * 0.75 s)
d3 = 2.0 cm * cos(3π/2)
d3 = 2.0 cm * 0
d3 = 0 cm

For t4 = 1.0 s:
d4 = 2.0 cm * cos(2π * 1.0 s)
d4 = 2.0 cm * cos(2π)
d4 = 2.0 cm * 1
d4 = 2.0 cm

Therefore, the positions of the object at the given times are:
d1 = 0 cm
d2 = -2.0 cm
d3 = 0 cm
d4 = 2.0 cm

If your answers still do not match what is expected by the system, make sure you are using the correct units (cm for position) and double-check your calculations.