A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and has vₓ = ─30 cm/s. Determine: h. The position at t = 0.40 s.

Video transcript

Hey, everyone. So this problem is dealing with simple harmonic motion. Let's see what it's asking us. We have a spring connected to a block of mass g. The block is oscillating horizontally at a frequency of four Hertz. If the initial position of the block is at X equals eight centimeters with a speed of 25 centimeters per second. What is the position of the block at time equals 0. seconds. Our multiple choice answers here are a 3.4 centimeters. B 8.0 centimeters C 9.4 centimeters or D 8.8 centimeters. So the first step to solving this problem is recalling that our angular frequency is given by the equation Omega equals two pi F where is our frequency given to us in the problem of four Hertz? So our angular frequency or there's system in the problem is a pi radiance per second. And then we can use our conservation of energy equations where we can write one half K A squared is equal to one half K X squared plus one half and B squared is equal to mass multiplied by the angular frequency squared. And so when we plug that in the mask, we're given in the problem of 350 g. So that's gonna be three, it's gonna be 30.350 kg multiplied by eight pi radiant per second. And that quantity is square. We plug that in and we get K our spring constant of 221 0.1 newtons per meter. And so we can look that in to our energy equation or conservation of energy equation to solve or a which is our maximum amplitude. And so that looks like one math multiplied by 221.1 m per meter multiply by A squared equals one half. Again, spring constant 221.1 newtons per meter X squared. So X was given to us as eight centimeters and the problem so we're gonna rewrite that is 0.8 m and that quantity squared plus one half times the mass 0. kg times our velocity square, the velocity is given as 25 centimeters per second. So again, keeping everything in standard units, we're gonna rewrite that as 250.25 m per second, not as swear. So when we, so for our amplitude, we get zero point 0806 m. So that's our maximum amplitude. And now we can use our simple harmonic motion equations to find that position. So we can recall that for simple harmonic motion position. Is given as X of T is equal to amplitude multiplied by the cosine of Omega T plus five. When X is zero or at time equals zero, our amplitude is going to be zero point m multiplied by the cosine of time of zero. So that term goes to zero. So it's just the cosine of five uh X equals but at the position at time equals zero, that is eight centimeters or 80.8 m. And then from there we can sorry, 0806 m. There we go, we can solve 45 and so five equals 0.1 22. And that unit is radiance. And so now we have everything we need to solve this equation again, except for, instead of time T equals zero, we're going to solve it for X at time equals 0. seconds. So that is again, our amplitude, 0.806 m multiplied by our cosine of O Omega we found was eight high radiant per second, multiplied by 0.2 seconds plus 0.122 ratings. And when we put that into our calculator, we get X equals 0.34 m or 3.4 centimeters. And so that is the correct answer for this problem. And that aligns with answer choice A so that's all we have for this one. We'll see you in the next video