Question

An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00 s the cart is released at the position x = +0.250 m. what is the position of the cart when t = 29.6

Answer 1

The cart is moving by simple harmonic motion, and its position at time t is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the oscillation
$\omega$ 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
$\omega = \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s$

Therefore now we can calculate the position of the system at the time t=29.6 s:
$x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m$

Answer 2

The position of air cart at $t=29.6{\text{ s}}$  is $\boxed{0.0460{\text{ m}}}$ .

Explanation:

It is given that the air track cart completes one oscillation for simple harmonic motion (SHM) in every $5.67{\text{ s}}$ .

Initially at time $t=0{\text{ s}}$  at position $x=0.25{\text{ m}}$  the cart is released.

Our aim is to obtain the position of air track cart for time $t=29.6{\text{ s}}$ .

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.

$f(t)=Acos(\omega t)$                    ......(1)

Here, $A$  is the amplitude for the maximum periodic function at time $t=0$  and ${{\omega }}$  is the angular frequency and  $t$ is time in seconds.

It is given that initially at time $t=0$  the position is $x=0.25$ , so the amplitude of the function A is $0.25$ .

The angular frequency is an integral multiple of $2{{\pi }}$  radians so, its value can be obtained as,

$\omega=\frac{2\pi}{T}$

Substitute $5.67$  for  $T$ in above equation to obtain the angular frequency as follows:

$\begin{aligned}\omega&=\frac{2\pi}{5.67}\\&=1.108\text{rad/s}\end{aligned}$

Substitute $0.25$  for $A$ , $1.108$  for ${{\omega }}$  and $29.6$  for $t$  in equation (1) to obtain the value of position.

$\begin{aligned}f\left(t\right)&=0.25{\text{ m}}\cos\left[{\left({1.108}\right)\left({29.6}\right){\text{rad}}}\right]\\&=0.25{\text{ m}}\cos\left({32.8}\right)\\&=0.25{\text{ m}}\left({0.1856}\right)\\&=0.0460{\text{ m}}\\\end{aligned}$

Therefore, the displacement at time $t=29.6{\text{ s}}$  is $0.0460{\text{ m}}$ .

Thus, the position of air cart at $t=29.6{\text{ s}}$  is $\boxed{0.0460{\text{ m}}}$ .

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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.