An air-track glider attached to a spring oscillates with a period of 1.5 s. At t = 0 s the glider is 5.00 cm left of the equilibrium position and moving to the right at 36.3 cm/s. a. What is the phase constant?

Video transcript

Hey, everyone. So this problem is working with simple harmonic motion. Let's see what it's asking us determine the phase constant for an object of mass 0.5 kg attached to a spring. The object is oscillating with a period of two seconds. Initially, the object is three centimeters to the left of its equilibrium position, excuse me, equilibrium position. And it has a velocity of 20 centimeters per second in the positive direction of the X axis. Our multiple choice answers here are a 1. radiance B 2.13 radiant C 1.13 radiant or D 4.27 radiant. So the key to solving this problem that's asking us to find this phase constant is we're calling our position and velocity equations which have that phase constant in them. So our position equation when we're dealing with simple harmonic motion is given by X of T is equal to a multiplied by cosine of omega T plus five where fine is that phase constant. And then our velocity of T is given by negative a multiplied by Omega multiplied by the sign of Omega T plus five. And so they give us our initial position and our initial velocity. So our X knot or initial position is going to be a multiplied by the cosine of five because T is zero. And similarly, our V knot is going to be negative a omega multiplied by the sign that's fine. And we are trying to solve for five, we know our X knot, our V knot. And if we, we can see that if we divide these two equations by each other, the amplitudes will cancel. So that is what we are going to do. Next. The knot divided by X knot is equal to negative A omega multiplied by the sign of divided by a multiplied by the cosine of five. And so like I said, the A is canceled and sine divide sine divided by cosine, we can recall is the same thing as the tangent. So we can rewrite this as P divided by X knot. We're going to move our omega and the negative sign. So we, we'll divide each of or sorry, we'll divide both sides by that. So we are left with just sine of five over cosine of five. And then that is the tangent of P five. And then the last step here before we solve and we take the inverse tangent is to recall that Omega is equal to two pi divided by T. And so now when we take that inverse tangent, we've got five is equal to the inverse tangent of negative V knot multiplied by T divided by X, not multiplied by two pi and they plug in our known values from the problem. So V was given to us as 20 centimeters per second. We're going to rewrite that in um to keep it in standard units. So 200.2 m per second, that was negative multiplied by hour period, which is two seconds divided by X knot. Again, given to us as three centimeters. Now, it's important to see here that it's three centimeters to the left. So that's gonna be a negative three centimeters for negative 0. m and then that's multiplied by two pi And so when we plug that into our calculator, we get 1.13 radiations or 4. ratings. And so this is the last kind of tricky part of this problem. We know from the problem that the object is in a negative position moving, right. Therefore, it is in the third quadrant, which means that the correct answer is 4. and not 1.13. And so when we look at our multiple choice answers that aligns with answer choice D, so that's all we have for this one, we'll see you in the next video.