A mass attached to a spring oscillates with a period of 3.15 s.
If the mass starts from rest at x = 0.0530 m and time t = 0, where is it at time t = 6.69 s?
I found online that you use the equation x = Acos(2pi*t/T) where A is the amplitude, t is 6.69s and T is 3.15s but I keep getting the wrong answer so I don't know if Im just plugging it in wrong or not. The answer I got was 0.05156m
Nevermind! I just forgot to change from degrees to radians on my calculator
Well done! Thanks for letting us know :)
To find the displacement of the mass at time t = 6.69 s, you can use the equation for simple harmonic motion:
x = A * cos(2πt / T)
Given:
T = 3.15 s (period)
A = 0.0530 m (amplitude)
t = 6.69 s (time)
Plugging in the values into the equation, we get:
x = 0.0530 * cos(2π * 6.69 / 3.15)
Calculating further, we have:
x = 0.0530 * cos(2.122)
Using a calculator or math software, we find:
x ≈ 0.0530 * (-0.174)
x ≈ -0.009222
Therefore, at time t = 6.69 s, the mass is located at approximately -0.009222 m.
To find the position of the mass at a given time using the equation x = Acos(2πt/T), you need to plug in the values correctly. Let's go through the steps:
Given:
Period, T = 3.15 s
Initial position, x = 0.0530 m
Time, t = 6.69 s
First, let's calculate the angular frequency (ω) using the formula ω = 2π/T:
ω = 2π / 3.15 = 1.996 rad/s
Now, we can plug in the values into the equation x = Acos(2πt/T):
x = Acos(1.996t/3.15)
To find the constant A (amplitude), we use the initial position, x = 0.0530 m:
0.0530 = A * cos(1.996 * 0 / 3.15)
0.0530 = A * cos(0)
Since cos(0) = 1, we can solve for A:
0.0530 = A * 1
A = 0.0530 m
Now, we can calculate the position at time t = 6.69 s:
x = 0.0530 * cos(1.996 * 6.69 / 3.15)
x ≈ 0.0428 m
Therefore, the correct position of the mass at t = 6.69 s is approximately 0.0428 m, not 0.05156 m as you calculated.