A mass hanging at the end of a spring oscillates up and down...

A mass hanging at the end of a spring oscillates up and down from its equilibrium position with velocity $$v(t)=3 \cos \left(\frac{3 t}{\sqrt{2}}\right)-3 \sqrt{2} \sin \left(\frac{3 t}{\sqrt{2}}\right)$$ centimeters per second, as shown in the figure. The mass is at its equilibrium at $t=0$. Use definite integrals to determine whether the mass will be above or below its equilibrium position at times $t=4$ and $t=5$.

A mass hanging at the end of a spring oscillates up and down from its equilibrium position with velocity
$$v(t)=3 \cos \left(\frac{3 t}{\sqrt{2}}\right)-3 \sqrt{2} \sin \left(\frac{3 t}{\sqrt{2}}\right)$$
centimeters per second, as shown in the figure. The mass is at its equilibrium at $t=0$. Use definite integrals to determine whether the mass will be above or below its equilibrium position at times $t=4$ and $t=5$.

Key Concepts

Interpreting the Sign of Displacement

After computing the displacement through definite integration, the sign of the result indicates the position relative to the equilibrium. A positive displacement typically shows that the mass is above equilibrium, whereas a negative displacement implies it is below equilibrium. This interpretation is key to answering questions regarding the location of the oscillating mass at given times.

Initial Conditions

Initial conditions, such as being at equilibrium at time zero, are crucial for determining the constants of integration when reconstructing the position function from the velocity function. They provide the necessary reference point to accurately assess the position of the mass at any later time.

Harmonic Oscillator

A mass-spring system exhibits harmonic oscillation, meaning its motion can be described by trigonometric functions such as sine and cosine. This periodic behavior is characteristic of systems where the restoring force is proportional to the displacement from the equilibrium, leading to simple harmonic motion.

Definite Integration

Definite integration is used to compute the net change of a quantity over an interval. In the context of motion, integrating the velocity function between two time points gives the total displacement, which can then be analyzed to determine if the object has moved above or below its starting point.

Relationship between Velocity and Displacement

In kinematics, the displacement of an object is obtained by integrating its velocity with respect to time. This fundamental relationship allows one to determine the position (or displacement) from the given velocity function through definite integration over a specified time interval.

Transcript

00:01 We wish to find the position function at t, given that the acceleration at t is sine pi t, the velocity at zero is three, and the position at zero is zero.

00:14 To find the velocity from the acceleration, we would integrate.

00:25 This is gonna fit the form of sine u du.

00:29 U is the pi t, the derivative of pi times the variable t is pi dt.

00:38 So what we have is the integral of sine u.

00:47 For the du, we have one dt, we needed pi dt.

00:51 What we have over what we need means we have one over pi of the du.

00:57 The integral of sine u du is negative cosine u.

01:03 So now i have negative one over pi cosine u, plus c, since it's indefinite, and since t is pi over two, i have negative one over pi cosine pi t plus c.

01:24 The c i can find because i know that the velocity is three when the time is zero.

01:31 So if i fill in zero for t, i can set the velocity equal to zero.

01:40 I'm sorry, set the velocity equal to three.

01:43 So if i do that, if i set t equal to zero, i should get an answer of three.

01:50 So that would give me negative one over pi times cosine zero plus c equals three.

02:02 The cosine of zero is actually one, so that would give me negative one over pi plus c equals three finally, if i add the one over pi over c, ends up being three plus one over pi.

02:20 So i actually have a velocity function that's negative one over pi times cosine pi t plus my c constant, which is three plus one over pi.

02:47 Finally, the position function is zero when the time is zero.

02:57 So let's integrate the velocity to get the position function.

03:06 I'm gonna break it into a couple integrals, one containing the cosine pi t, and the other containing the constants, three plus one over pi.

03:21 The cosine pi t integral will fit a u du again.

03:30 U is pi over t, du is pi dt.

03:37 So that first integral becomes the integral of cosine u.

03:47 For the du, we have negative one over pi, but du is pi dt...