A weight on the end of a coiled spring is pulled
10cm
below its resting position. The weight is released at time
=t0
seconds and moves upward, oscillating in simple harmonic motion. It completes one cycle in
3
seconds. (Note that upward is the positive direction.)
Give the equation modeling the displacement
d
as a function of time
t
.
see related questions below
again
d = 8 sin (20 π t)
To model the displacement of the weight as a function of time, we can use the equation for simple harmonic motion:
d = A * cos(ωt + φ)
Where:
- d is the displacement (in this case, it will be positive for upward motion and negative for downward motion)
- A is the amplitude (the maximum displacement from the resting position, which is 10cm in this case)
- ω is the angular frequency (related to the time period by ω = 2π / T, where T is the time period, which is 3 seconds in this case)
- t is the time
- φ is the phase constant (the initial phase of the motion, which we need to determine)
To determine the value of φ, we need to consider the initial conditions of the motion. In this case, the weight is pulled 10cm below its resting position, so when t = t0 (at time of release), d = -10cm.
Substituting these values into the equation, we have:
-10 = A * cos(ωt0 + φ)
Now we need to solve for φ.
To do that, divide both sides of the equation by A:
-10 / A = cos(ωt0 + φ)
Now, take the inverse cosine of both sides:
cos^(-1)(-10 / A) = ωt0 + φ
Finally, solve for φ:
φ = cos^(-1)(-10 / A) - ωt0
Substituting the values of A, ω, and t0 as given in the problem, we can calculate φ:
φ = cos^(-1)(-10 / 10) - (2π / 3)t0
Simplifying:
φ = cos^(-1)(-1) - (2π / 3)t0
Since cos(π) = -1, we have:
φ = π - (2π / 3)t0
Now we can use this value of φ to write the equation modeling the displacement as a function of time:
d = 10 * cos(ωt + π - (2π / 3)t0)