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: e. The maximum speed.

Video transcript

Hey, everyone. So this problem is dealing with conservation and energy and simple harmonic motion. Let's see what it's asking us. An object of mass 300 g is attached to a horizontal spring. It is experiencing oscillations with a frequency of three Hertz determine the maximum speed of oscillations. If the object's initial position is X equals 10 centimeters and the, the velocity is V equals 20 centimeters per second. Our multiple choice answers here are a 18.8 m per second. B one m per second. C 1.9 m per second or D 10.5 m per second. OK. So the way we were going to address this problem is they're asking us to find maximum speed. So we can first recall that with angular momentum oscillations like this, our maximum speed is given by Omega annular frequency multiplied by a the amplitude. And in turn, that omega at angular frequency is given by two pi F. We are asked to solve for V max and we can pretty easily find omega, but it's going to take us another extra couple of steps to find a our amplitude. So let's just solve through for Omega here, you had two pi multiplied by F or frequency was given the problem as three Hertz. And so our angular frequency is 65 and those units are radiant per second. Now, for a, we can recall that our conservation of energy equation when we're working with springs looks like one half K A squared or K is our spring constant equals one half K X squared plus one half and V squared. And in turn K is given by and multiplied by omega squared. So when we pair those two equations, we can see that we now have everything we need to solve for a, our amplitude. And then we'll plug that back in to find our maximum velocity. OK. So K, our spring constant is simply mass multiplied by our angular frequency squared mass is given to us in the problem 300 g, we're going to keep the unit standard. So 300 g is the same as 3000.3 kg. Then our angular frequency is six pi gradients per second. And that quantity is squared, plug that into our calculator and we get a spring constant of 106.6 newtons per meter. Now, back to our conservation energy equation, you can see we had a half in every term to those cancel, we're solving for a. So rearranging and isolating that variable A is equal to the square root of K X squared plus M B squared, all divided by K and we now have everything we need to solve for this problem. So let's plug in those known values. OK. We um sorry K we just saw for 106.6 newtons per meter, our initial position X was given to us as 10 centimeters. Again, I'm going to rewrite that as 100.1 m to keep us in standard units plus the mass up 0.3 kg multiplied by the velocity squared. So that velocity was given to us as 20 centimeters per second. And again, in standard units, that's 0.2 m per second. OK. I'm sorry, I'm missing a square here. That's gonna be K X square. And then all of that divided by K. So again, 106.6 newtons per meter, when we solve that out for a, we get 0.1 m. And so then our last step is to go back to our V max equation where we have our angular frequency six pi meter, six I gradients per second multiply by hour, amplitude, 0.1 m and that equals 1.9 m per second. So that is the answer here and that aligns with cancer choice C. So that's all we have for this one. We'll see you in the next video.