Respuesta :
The cart is moving by simple harmonic motion, and its position at time t is described by
[tex]x(t) = A \cos (\omega t)[/tex]
where
A is the amplitude of the oscillation
[tex]\omega[/tex] is the angular frequency
The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released:
A=0.250 m
The period of the motion is T=5.67 s, and the angular frequency is related to the period by
[tex]\omega = \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s [/tex]
Therefore now we can calculate the position of the system at the time t=29.6 s:
[tex]x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m[/tex]
[tex]x(t) = A \cos (\omega t)[/tex]
where
A is the amplitude of the oscillation
[tex]\omega[/tex] is the angular frequency
The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released:
A=0.250 m
The period of the motion is T=5.67 s, and the angular frequency is related to the period by
[tex]\omega = \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s [/tex]
Therefore now we can calculate the position of the system at the time t=29.6 s:
[tex]x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m[/tex]
The position of air cart at [tex]t=29.6{\text{ s}}[/tex] is [tex]\boxed{0.0460{\text{ m}}}[/tex] .
Explanation:
It is given that the air track cart completes one oscillation for simple harmonic motion (SHM) in every [tex]5.67{\text{ s}}[/tex] .
Initially at time [tex]t=0{\text{ s}}[/tex] at position [tex]x=0.25{\text{ m}}[/tex] the cart is released.
Our aim is to obtain the position of air track cart for time [tex]t=29.6{\text{ s}}[/tex] .
The function of time can be called as displacement. Since, this is a periodic motion the function of time will be periodic in nature.
The expression for simple periodic function is shown below.
[tex]f(t)=Acos(\omega t)[/tex] ......(1)
Here, [tex]A[/tex] is the amplitude for the maximum periodic function at time [tex]t=0[/tex] and [tex]{{\omega }}[/tex] is the angular frequency and [tex]t[/tex] is time in seconds.
It is given that initially at time [tex]t=0[/tex] the position is [tex]x=0.25[/tex] , so the amplitude of the function A is [tex]0.25[/tex] .
The angular frequency is an integral multiple of [tex]2{{\pi }}[/tex] radians so, its value can be obtained as,
[tex]\omega=\frac{2\pi}{T}[/tex]
Substitute [tex]5.67[/tex] for [tex]T[/tex] in above equation to obtain the angular frequency as follows:
[tex]\begin{aligned}\omega&=\frac{2\pi}{5.67}\\&=1.108\text{rad/s}\end{aligned}[/tex]
Substitute [tex]0.25[/tex] for [tex]A[/tex] , [tex]1.108[/tex] for [tex]{{\omega }}[/tex] and [tex]29.6[/tex] for [tex]t[/tex] in equation (1) to obtain the value of position.
[tex]\begin{aligned}f\left(t\right)&=0.25{\text{ m}}\cos\left[{\left({1.108}\right)\left({29.6}\right){\text{rad}}}\right]\\&=0.25{\text{ m}}\cos\left({32.8}\right)\\&=0.25{\text{ m}}\left({0.1856}\right)\\&=0.0460{\text{ m}}\\\end{aligned}[/tex]
Therefore, the displacement at time [tex]t=29.6{\text{ s}}[/tex] is [tex]0.0460{\text{ m}}[/tex] .
Thus, the position of air cart at [tex]t=29.6{\text{ s}}[/tex] is [tex]\boxed{0.0460{\text{ m}}}[/tex] .
Learn More:
1. Waves https://brainly.com/question/3293068
2. Friction https://brainly.com/question/11746789
3. Energy https://brainly.com/question/8577648
Answer Details:
Grade: High School
Subject: Physics
Chapter: Oscillations
Keywords:
Air track, cart, spring, attached, oscillation, simple, harmonic, motion, position, released, time period, angular frequency, argument, periodic, displacement, amplitude, SHM.
