Scientists are measuring the properties of a newly discovered elastic material. They create a 1.5-m-long, 1.6-mm-diameter cord, attach an 850 g mass to the lower end, then pull the mass down 2.5 mm and release it. Their high-speed video camera records 36 oscillations in 2.0 s. What is Young's modulus of the material?

Video transcript

Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A group of researchers announced that the development of a new rubber like material with surprising properties. A bungee jumper receives a 2.0 m long rope with a 1.0 millimeter radius made of this rubber material. He connects a 10 kg mass to the end of the rope and suspends it vertically by attaching the other end to a tall structure from its equilibrium position. He takes the mass down 10 centimeters and lets it go using a motion detector. He counts 15 oscillations in 4.5 seconds. Calculate the rubber young's modulus and that is our end goal. OK. So we're given some multiple choice answers. They're all in the same units of newtons per meters squared. So let's read them off to see what our final answer might be. A is 1.2 multiplied by 10 to the eighth. Power B is 1.6 multiplied by 10 to the eighth. Power C is 2.8 multiplied by 10 to the ninth power and D is 3.2 multiplied by 10 to the ninth power. Ok. So first off, let us recall that the young's modulus describes a rubber band material that can be easily stretched and deformed. Also recall and use capital Y to denote the young's modulus. So once we can like, so now we need to recall and use Young's modulus equation which states that why is equal to the force applied divided by the cross sectional area of the rope capital A multiplied by delta L divided by elf. So delta L divided by L is the change in the length of the rope. OK. So let us model the rope as a spring and the mass attached as a particle undergoing a simple harmonic motion. And let us ignore the dampening effect if a mass M is attached to an ideal spring, which in this case, we assume the rope is acting as an ideal spring. So we need to focus on a spring constant K. So it's an ideal spring with a spring constant K. Thus, we can write in what's called the Young's modulus equation one. So let's call equation two F. The four supplied is equal to the spring constant multiplied by delta L. So now we can combine equations one and two and let's call it equation three. So the equation three states that Y is equal to K multiplied by L all divided by a. So the spring constant multiplied by the length divided by the cross sectional area of the rope. So the mass will oscillate with a period that is written as capital T represents the period is equal to two pi multiplied by the square root of the mass divided by the spring constant. So now we need to rearrange this equation to solve for K using a little bit of algebra. So when we do that to isolate K, we get K is equal to four pi squared. So four multiplied by pi squared multiplied by the mass divided by the period squared. So mass divided by the period squared. So the period of motion can be found by dividing the time needed to complete 15 oscillations by 15. So let's do that really fast. So T is equal to 4. seconds divided by 15 is equal to 0.3 seconds. Now we can substitute this value back into our equation to solve for the spring constant. So K is equal to four pi so four multiplied by pi squared multiplied by the mass which was kg divided by the period, which the period was 0.3 seconds squared, which is equal to. So once we plug that into a calculator, we should get that K is equal to 386.5 newtons per meter. OK. So, OK, so now we can use this value to help us solve for A Y, which is our end goal to find our final answer by plugging it back into equation three. So let's do that. So why is equal to 4386. newtons per meter multiplied by 2 m. This is the length of the rope 2 m divided by the cross sectional area. So let's note that the cross sectional area is gonna represent the area of a circle. So the area of a circle is a, is equal to pi multiplied by the radius square. So we need to convert our radius from millimeters to meters. So let's do that as a quick dimensional analysis problem. So it's gonna be pi multiplied by one millimeter multiplied. So in 1 m, there is 1000 millimeters. So the millimeters cancel out leaving us with meters. So that means that our meter amount is 0.001 m squared. OK. So just to make it simple, it's gonna be pi multiplied by 0.001 m square. So when we plug into a calculator, we should get that Y is equal to 2. multiplied by 10 to the power of nine newtons per meters squared. Fantastic. That's our final answer. We did it. So let's go look at the multiple choice answers to see what our final answer is. So the correct answer is the letter C 2.8 multiplied by 10. The power of nine newtons per meter squared. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.