How to of the Day

---

• How to Do a Kung Fu Style Full Body Workout
• How to Make Stuffed Meatballs
• How to Solve the Pendulum

How to Do a Kung Fu Style Full Body Workout Posted: 23 Jun 2017 05:00 PM PDT A kung fu-style workout is a great way to work your whole body. Like any workout, you need to start by warming up. With kung fu, you need to start with exercises that will warm up all your muscles and get your blood pumping. Then, you can move on to working both your upper and lower body. You can actually alternate between upper and lower body exercises, and it doesn't hurt to throw in some jumping jacks or other exercises between movements in your workout. EditSteps EditWarming Up Do some jumping jacks. One effective warm-up is the standard jumping jacks that you did in elementary school. Stand with your arms at your side and your legs together. Simultaneously jump your legs out to shoulder-width apart as you bring your arms straight out from your shoulders and then straight up above your head.[1] Do a set amount, such as 20 jumping jacks or 20 seconds of jumping jacks. Work up to more as you can. Add squat jumps. Begin with a squat, where you act like you're sitting in a chair but there's no chair behind you. You'll need your feet shoulder-width apart. Slowly lower yourself down like you're sitting. You should reach a point where your knees are at a 90-degree angle.[2] Now, simultaneously raise your arms above your head as you jump into the air from the squat position. Land in a standing position, and do it all again.[3] Try starting with 5 reps and moving up to 10 gradually. Jump rope. Jumping rope is another good warm-up that may take you back to your grade-school days. Get a sturdy jump rope, and try jumping continuously for a set period. It gets your heart pumping and warms up muscles throughout your body.[4] Do what you can at first, and work up to jumping for 5 minutes. Make quick position changes. One way to really prepare you for your workout ahead is to try some quick position changes that you'd normally need to do to move around in kung fu. For example, with your left forward, spring lightly forward and back a few times, then switch quickly to the right foot forward.[5] Punch the air. Since you'll use punch-like movements in your workout, do some punches while warming up. Start by punching with one arm only, ten times in a row. After you've completed that move, switch to the other arm.[6] Start with your left foot a bit forward. Spring forward a bit on both feet, then punch forward with your left arm. It's similar to a boxing move. Do a series of jumping oblique twists. This movement warms and stretches your whole body. Start with your feet together. Your arms should be in front of your chest, with your elbows out and palms towards the ground. Start by jumping in the air and twisting your feet and knees to the right, then jumping again and twisting to the left in quick succession. Your chest should stay forward. Keep going back and forth quickly.[7] Try 30 seconds of this move at a time. Keep moving. If you need a break between your warm-up exercises, that's fine. However, you should still keep moving. When you're taking a break, you need to jog in place to keep the momentum going. Once you've recovered a bit, try another warm-up exercise.[8] Make sure you're rotating through exercises. That is, you can do jumping jacks once, then go back to them after doing a couple of other exercises. Try the "world greatest stretch." This yoga stretch works to stretch out your whole body, which is essential when doing a kung fu workout. Start with a forward lunge. Step far forward with one foot. As you do, lower yourself towards the ground. The front knee should reach a 90-degree angle and the back knee should almost touch the ground. Hold this move for about 10 seconds.[9] Using the arm that's on the same side as your forward leg, bend it at the elbow and lean as far forward as you can on the inside of your leg. You can try to the touch the ground with the elbow. If you can't touch the ground, just get as far as you can. Your other hand should be flat on the ground to help you stay upright. Hold for about 10 seconds. Next, place your hands on both sides of your foot. You may need to support yourself on your fingers. Straighten your front leg up, moving the back leg as needed, and lift up your front foot's toes. Hold for 10 seconds. Move to put the other leg forward, and repeat. EditWorking the Upper Body Do an upward block. Make a fist out of your hand. Your arm should be bent. Bring that arm out in front of you with your forearm facing out. Your arm should be about waist level and parallel to the ground. Now, raise the arm up, bringing it in front of your face and then up above your head. Your arm should now be just above your forehead with the forearm still out. Bring your arm back to the start.[10] Alternate arms for 20 reps. Start out slow, and work up to a faster pace over time. You can add more blocks as you get stronger. Move to downward blocks. Begin in horse stance. Make fists with both hands, and bend your arms at the elbow. The inner arm should be facing upwards. Move one arm out in front, flattening out to an open hand, facing downward. Your arm should be at about waist-height.[11] Move your arm down, pressing a bit harder with the outside of the wrist as you hit the "bottom" of the movement. Move back to starting position. Alternate between hands for about 20 reps. You can move up to doing more and doing them faster as you get stronger. For horse stance, place your feet a bit wider than your hips. Your toes should be pointed outwards. Keeping your back straight, bend at the knees until they are just over your toes.[12] Alternate punches. Start in a high horse stance, meaning don't go as deep as you did in the downward blocks. You arms should be bent at the elbow and by your side, with the underarms facing upwards and fists clenched. Start by punching one arm forward.[13] As you punch forward, rotate the wrist so your inner arm is facing down by the time your arm is extended. As you come back, rotate it back upwards. Punch straight out from the center of your body, rotating your torso back and forth to punch. Move back and forth between arms for about 30 seconds. To make it a bit more difficult, get lower in the horse stance. Try push-ups with extra stretches. Begin in a push-up position. Lay face down on the ground with your toes holding your lower end up (you can also use your knees instead). Place your palms on the ground (or knuckles, which is harder). Lower yourself to the ground, then bring yourself back up, keeping your body straight throughout.[14] Try ten reps, then stop. Using one arm to balance you in the center, stretch the other arm out straight up from your shoulder. Hold for ten seconds. Next, jump your legs out, and turn your body towards the arm you have out, including turning your feet. Stretch that arm up into the air. Your face should be looking upward. Hold for ten seconds. Repeat the whole movement on the other side. Do another five pushups. EditWorking the Lower Body Perform straight kicks. Start in bow stance by having one leg in front and one in back. You'll be kicking with the one in back. Place your hands on your hips in preparation for the move. Rock forward on to you front leg a bit to begin the move.[15] As you move your weight to your front leg, your back leg should begin to lift off the ground. Keep it straight. Use the muscles in that leg to kick it forward, while pushing up with the other leg, keeping that knee slightly bent, not locked. Kick as high as you can and still keep your balance. Use your muscles to pull the leg back down to the starting position. Try 20 kicks on one side, aiming for waist-high, then move to the other leg and repeat. Do round kicks. Begin in a defensive bow stance, with one leg in front of the other. You can bring your arms up in front of you like you're getting ready to punch. Make fists and have your underarms facing your body with your elbows bent.[16] Shift your weight to your front leg, lifting the back leg off the ground. Instead of going straight forward, lift it to the outside and bring it forward. As you bring it forward, move it from outside to inside. Basically, you are lifting it and swiveling your foot so it comes up with the toe pointing to the other side of your body as the foot moves in front of you. Your knee should be bent until your leg gets in front of you, then you should kick out. Your foot on the ground will turn, and your body will lean slightly way from the kick. Do ten to fifteen reps and move to the other side. Work on crescent kicks. Start out in the bow position again. Bring your back leg forward, moving it a bit in front of your other leg as you bring it into the air. Now, your foot should go in an arch in front of your body. As you bring your leg in front of you back towards the other side, the knee should be facing your body and the sole of your foot should be pointing upwards. Bring it down the other side and back behind your body.[17] Try ten reps, and then move to the other leg. Try leg raises. Lay on your back. Your hands can be palms together on your chest, or you can use them under you to support your back/buttocks. Raise both legs into the air together, keeping the knees slightly bent. Lower them back towards the ground, but don't touch the ground.[18] Do ten reps with your legs together, then try scissoring your legs back and forth. EditSources and Citations Cite error: <ref> tags exist, but no <references/> tag was found

How to Make Stuffed Meatballs Posted: 23 Jun 2017 09:00 AM PDT Homemade meatballs make a delicious meal, but if you'd like to make them extra special, stuff them with rich cheese. Combine a basic meatball mixture using ground beef or a combination of beef and lamb. Wrap the meatball mixture around your favorite cheese (like mozzarella), so the cheese is encased in meat. You can then fry the meatballs for a quick and crispy dish or bake them for tender meatballs. Serve the stuffed meatballs immediately, so the gooey cheese can ooze out of them. EditIngredients EditFor Baked Stuffed Meatballs Makes 20 stuffed meatballs 1 pound (454 g) of ground beef 3/4 cup (70 g) of plain bread crumbs 1/2 cup (50 g) of grated Parmesan cheese 1/2 cup (120 ml) of water 2 tablespoons of chopped fresh parsley 1 egg 1 1/2 teaspoons of garlic powder 1 teaspoon of salt 3/4 teaspoon of black pepper 1 (8-ounce or 226 g) block of mozzarella, cut into 20 (1/2-inch or 12 mm) cubes EditFor Fried Stuffed Meatballs Makes 23 stuffed meatballs 3/4 pound (350 g) of lean ground beef 2/3 cup (150 g) of ground lamb 1 onion, grated ½ teaspoon of salt ½ teaspoon of ground black pepper ¼ teaspoon of ground cumin 1 1/3 cups (150 g) of shredded cheese (like mozzarella) 2 tablespoons of olive oil to fry 2 tablespoons of fresh parsley, minced EditSteps EditMaking Baked Stuffed Meatballs Preheat the oven and cut the cheese. Turn on the oven to 350 degrees F (180 C). Take one 8-ounce (226 g) block of mozzarella cheese and use a sharp knife to cut it into 1/2-inch (12 mm) cubes. You should get about 20 small cubes of cheese.[1] You could also use pepper jack, cheddar, or monterey jack cheese. Combine the meatball ingredients and prepare a baking sheet. Spray a baking sheet with cooking spray and set it aside. Get out a large mixing bowl and place all of the meatball ingredients in it. Use a wooden spoon to stir the mixture well. You'll need:[2] 1 pound (454 g) of ground beef 3/4 cup (70 g) of plain bread crumbs 1/2 cup (50 g) of grated Parmesan cheese 1/2 cup (120 ml) of water 2 tablespoons of chopped fresh parsley 1 egg 1 1/2 teaspoons of garlic powder 1 teaspoon of salt 3/4 teaspoon of black pepper Form the stuffed meatballs. Gather a few tablespoons of meatball mixture into a ball and press one cheese cube in the center of it. Spread the meatball mixture around the cheese, so it is encased in meat. Continue to form the stuffed meatballs until you use up the meat mixture. You should have about 20 stuffed meatballs. Place them on the prepared baking sheet.[3] You can use spoons or a small cookie scoop to portion out the meatballs. Bake the stuffed meatballs. Place the baking sheet in the preheated oven and bake the stuffed meatballs for 15 to 20 minutes. The meat shouldn't be pink and should be at least 160 degrees F (70 C). Serve the stuffed meatballs right away.[4] Don't worry if some of the cheese cooks out of the meatballs, just try to remove them from the pan immediately, so they don't stick. EditMaking Fried Stuffed Meatballs Make and freeze shredded cheese balls. Get out 1 1/3 cups (150 g) of shredded cheese and use your fingers to pinch together a few teaspoons of cheese. Try to form the cheese into a ball shape. You should get about 20 to 23 small cheese balls. Place these on a baking sheet and freeze them for 30 minutes.[5] You can use Mozzarella, Colby, or Monterey Jack Cheese. Grate the onion and combine it with the meat. Peel one onion and grate it against the large-hole side of a box grater. Place the grated onion in a mixing bowl along with 3/4 pound (350 g) of lean ground beef and 2/3 cup (150 g) of ground lamb. Use a wooden spoon or your fingers to gently combine the onion and the meat.[6] If you can't find ground lamb, you can easily substitute ground pork. Season and chill the meatball mixture. Sprinkle the meatball mixture with ½ teaspoon of salt, ½ teaspoon of ground black pepper, and ¼ teaspoon of ground cumin. Stir the seasonings into the meat until they're incorporated. Place the mixture in the refrigerator and chill it for 30 minutes.[7] Chilling the meat mixture will help the flavors develop. Form the stuffed meatballs. Remove the meat mixture from the fridge and use a few spoons or a small cookie scoop to portion out meatballs. You should get between 20 and 23 meatballs. Gently flatten each meatball between the palms of your hand and put a frozen cheese ball in the center. Spread the meat over the cheese and form a round ball. Set each stuffed meatball on the baking sheet.[8] If the meat sticks to your hand, you can dip your hands in warm water. This will make it easier to handle the meat.[9] Fry the meatballs. Pour 2 tablespoons of olive oil into a large skillet and heat it over medium-high. Once the oil shimmers, add enough meatballs to cover the skillet in a single layer. Fry the meatballs for 10 minutes and use tongs to turn them every once in a while. The meatballs should be completely golden brown on the outside. Remove the meatballs to a serving plate and fry another batch of meatballs (if you couldn't fit all of them in at once).[10] You can garnish the stuffed meatballs with 2 tablespoons of minced fresh parsley. Consider serving the meatballs with a dipping sauce or glaze them with a little honey and pomegranate molasses. Finished. EditThings You'll Need Measuring cups and spoons Mixing bowl Wooden spoon Baking sheet Knife and cutting board Spoons or small cookie scoop Baking sheet Box grater Large skillet Tongs EditSources and Citations Cite error: <ref> tags exist, but no <references/> tag was found

How to Solve the Pendulum Posted: 23 Jun 2017 01:00 AM PDT A pendulum is an object consisting of a mass suspended from a pivot so that it can swing freely. The mathematics of pendulums are governed by the differential equation ${\displaystyle {\frac {{\mathrm {d} }^{2}\theta }{{\mathrm {d} }t^{2}}}+{\frac {g}{l}}\sin \theta =0}$ which is a nonlinear equation in ${\displaystyle \theta .}$ Here, ${\displaystyle g\approx 9.8\ {\text{m}}\,{\text{s}}^{-2}}$ is the gravitational acceleration, and ${\displaystyle l}$ is the length of the pendulum. Simple pendulums can be used to measure the local gravitational acceleration to within 3 or 4 significant figures. EditSteps EditSmall Angle Approximation Make the small-angle approximation. The governing differential equation for a simple pendulum is nonlinear because of the ${\displaystyle \sin \theta }$ term. In general, nonlinear differential equations do not have solutions that can be written in terms of elementary functions, and this is no exception. However, if we assume that the angle of oscillation is small, e.g. ${\displaystyle \theta <15^{\circ },}$ then it is reasonable to make the approximation that ${\displaystyle \sin \theta \approx \theta .}$ We see that ${\displaystyle \theta }$ is the first term in the Taylor series for ${\displaystyle \sin \theta }$ about ${\displaystyle \theta =0,}$ so our error in this approximation is on the order of ${\displaystyle O(\theta ^{3}).}$ ${\displaystyle \sin \theta =\theta -{\frac {\theta ^{3}}{3!}}+{\frac {\theta ^{5}}{5!}}-{\frac {\theta ^{7}}{7!}}+\cdots }$ We then obtain the equation for a simple harmonic oscillator. This equation is linear and has a well-known solution. ${\displaystyle {\frac {{\mathrm {d} }^{2}\theta }{{\mathrm {d} }t^{2}}}+{\frac {g}{l}}\theta =0}$ Solve the differential equation using the small-angle approximation. Since this is a linear differential equation with constant coefficients, our solution must either be in the form of exponentials or trigonometric functions. For physical reasons, we expect that the equation of motion be oscillatory (trigonometric) in nature. Obtain the characteristic equation and solve for the roots. ${\displaystyle r^{2}+{\frac {g}{l}}=0}$ ${\displaystyle r=\pm {\sqrt {\frac {g}{l}}}i}$ Since our roots are imaginary, our solution is indeed oscillatory, as expected. From the theory of differential equations, we obtain our solution below. We write the angular frequency ${\displaystyle \omega ={\sqrt {\frac {g}{l}}}.}$ ${\displaystyle \theta (t)=c_{1}\cos \omega t+c_{2}\sin \omega t}$ Write the equation of motion in terms of amplitude and phase factor. A more useful formulation of the solution involves making the following manipulation. Multiply the solution by ${\displaystyle {\frac {\sqrt {c_{1}^{2}+c_{2}^{2}}}{\sqrt {c_{1}^{2}+c_{2}^{2}}}}.}$ ${\displaystyle {\sqrt {c_{1}^{2}+c_{2}^{2}}}\left({\frac {c_{1}}{\sqrt {c_{1}^{2}+c_{2}^{2}}}}\cos \omega t+{\frac {c_{2}}{\sqrt {c_{1}^{2}+c_{2}^{2}}}}\sin \omega t\right)}$ Draw a right triangle with angle ${\displaystyle \phi ,}$ hypotenuse length ${\displaystyle {\sqrt {c_{1}^{2}+c_{2}^{2}}},}$ opposite side length ${\displaystyle c_{1},}$ and adjacent side length ${\displaystyle c_{2}.}$ Replace the constant ${\displaystyle {\sqrt {c_{1}^{2}+c_{2}^{2}}}}$ with a new constant ${\displaystyle \Theta ,}$ denoting amplitude. Now we can simplify the quantities in parentheses. The result is that the second arbitrary constant has been replaced with an angle. ${\displaystyle {\begin{aligned}\theta (t)&=\Theta (\sin \phi \cos \omega t+\cos \phi \sin \omega t)\\&=\Theta \sin(\omega t+\phi )\end{aligned}}}$ Because ${\displaystyle \phi }$ is arbitrary, we can also use the cosine function as well. Mathematically, the two phase factors are different, but in terms of finding the equation of motion given initial conditions, only the form of the solution matters. Writing it in terms of cosine is slightly more common because it fits initial conditions well (imagine a pendulum being let go at some angle - the cosine function fits this situation naturally). ${\displaystyle \theta (t)=\Theta \cos(\omega t+\phi )}$ Solve for initial conditions. Initial conditions are solved in the usual manner with regards to second-order differential equations given the general solution. Assume initial conditions ${\displaystyle \theta (0)=\theta _{0}}$ and ${\displaystyle {\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}(0)=0.}$ This is equivalent to saying that we release a pendulum without any force at some angle ${\displaystyle \theta _{0}}$ from equilibrium, provided that ${\displaystyle \theta _{0}}$ is not too great. Substitute these conditions into the general solution. Differentiate the general solution and substitute these conditions into that as well. We immediately obtain ${\displaystyle \phi =0}$ and ${\displaystyle \Theta =\theta _{0}.}$ ${\displaystyle \theta (t)=\theta _{0}\cos \omega t}$ If you are given numbers, then simply follow the above steps with the appropriate numbers substituted. Find the period of a simple pendulum. Physically, the angular frequency is the number of radians rotated per unit time. It is therefore related to the period via the relation ${\displaystyle \omega ={\frac {2\pi }{T}}.}$ We can then solve for the period ${\displaystyle T.}$ ${\displaystyle \omega ={\sqrt {\frac {g}{l}}}}$ ${\displaystyle T=2\pi {\sqrt {\frac {l}{g}}}}$ The order of ${\displaystyle g}$ and ${\displaystyle l}$ can get confusing. If it does, we go back to physical intuition. Intuitively, a longer pendulum should have a longer period than a shorter pendulum, so ${\displaystyle l}$ should be on top. EditArbitrary Angle Write the differential equation of a pendulum without the small-angle approximation. This equation is no longer linear and is not easily solved. It turns out that the period of such a pendulum can be written exactly in terms of elliptic integrals - integrals that historically were studied to find the arc length of ellipses, but naturally arise in the study of pendulums as well. ${\displaystyle {\frac {{\mathrm {d} }^{2}\theta }{{\mathrm {d} }t^{2}}}+{\frac {g}{l}}\sin \theta =0}$ To make things simple, we are given the same initial conditions as before: ${\displaystyle \theta (0)=\theta _{0}}$ and ${\displaystyle {\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}(0)=0.}$ Multiply the equation by ${\displaystyle 2{\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}}$. ${\displaystyle 2{\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}{\frac {{\mathrm {d} }^{2}\theta }{{\mathrm {d} }t^{2}}}+{\frac {2g}{l}}{\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}\sin \theta =0}$ We can then make use of the chain rule for both terms. ${\displaystyle 2{\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}{\frac {{\mathrm {d} }^{2}\theta }{{\mathrm {d} }t^{2}}}={\frac {\mathrm {d} }{{\mathrm {d} }t}}\left({\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}\right)^{2}}$ ${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }t}}\cos \theta =-\sin \theta {\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}}$ Then, we arrive at the following equation. ${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }t}}\left({\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}\right)^{2}={\frac {2g}{l}}{\frac {\mathrm {d} }{{\mathrm {d} }t}}\cos \theta }$ Integrate with respect to time. The integration introduces an integration constant. Physically, this constant represents the cosine of an initial angle. There are two solutions because the pendulum can move counterclockwise or clockwise. ${\displaystyle \left({\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}\right)^{2}={\frac {2g}{l}}(\cos \theta -\cos \theta _{0})}$ ${\displaystyle {\frac {{\mathrm {d} }\theta }{{\mathrm {d} }t}}=\pm {\sqrt {\frac {2g}{l}}}{\sqrt {\cos \theta -\cos \theta _{0}}}}$ Set up the integral to find the period. From our previous results, we found that ${\displaystyle \Theta =\theta _{0}}$ was the amplitude of oscillation. This suggests that half the period is the time taken for the pendulum to traverse from ${\displaystyle -\theta _{0}}$ to ${\displaystyle \theta _{0}.}$ ${\displaystyle {\frac {T}{2}}={\sqrt {\frac {l}{2g}}}\int _{-\theta _{0}}^{\theta _{0}}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta -\cos \theta _{0}}}}}$ Because ${\displaystyle \cos \theta }$ is even, we can factor out a 2. ${\displaystyle T=2{\sqrt {\frac {2l}{g}}}\int _{0}^{\theta _{0}}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta -\cos \theta _{0}}}}}$ This integral is tough, and cannot be evaluated using elementary methods. However, it can be evaluated exactly in terms of the Beta function if we assume that ${\displaystyle \theta _{0}=\pi /2,}$ i.e. the angle of oscillation is 90°. This is large enough to be outside the scope of the small-angle approximation. We do this calculation in the next step. Solve for the period given an oscillation angle of 90°. When ${\displaystyle \theta _{0}=\pi /2,}$ ${\displaystyle \cos \theta _{0}=0}$ and we obtain the following integral. ${\displaystyle T=2{\sqrt {\frac {2l}{g}}}\int _{0}^{\pi /2}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta }}}}$ This integral still does not have an antiderivative that can be written in terms of elementary functions, but it can be evaluated exactly in terms of the Beta function, itself written in terms of the Gamma function. ${\displaystyle {\frac {\Gamma (\alpha )\Gamma (\beta )}{2\,\Gamma (\alpha +\beta )}}=\int _{0}^{\pi /2}\cos ^{2\alpha -1}\theta \sin ^{2\beta -1}\theta {\mathrm {d} }\theta }$ We see from direct comparison that ${\displaystyle \alpha =1/4}$ and ${\displaystyle \beta =1/2.}$ Given that ${\displaystyle \Gamma (1/2)={\sqrt {\pi }},}$ we arrive at the following answer. ${\displaystyle \int _{0}^{\pi /2}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta }}}={\frac {\sqrt {\pi }}{2}}{\frac {\Gamma (1/4)}{\Gamma (3/4)}}}$ We now make use of Euler's reflection formula to simplify, since ${\displaystyle \Gamma (3/4)}$ is related to ${\displaystyle \Gamma (1/4).}$ ${\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin(\pi z)}}}$ ${\displaystyle \int _{0}^{\pi /2}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta }}}={\frac {\Gamma ^{2}(1/4)}{2{\sqrt {2\pi }}}}}$ Combining with our previous result, and setting the period of the pendulum with small-angle approximation ${\displaystyle T_{0},}$ we arrive at the following result. Note that ${\displaystyle \Gamma (1/4)}$ is transcendental. ${\displaystyle T={\sqrt {\frac {l}{g}}}{\frac {\Gamma ^{2}(1/4)}{\sqrt {\pi }}}={\frac {\Gamma ^{2}(1/4)}{2\pi ^{3/2}}}T_{0}\approx 1.18T_{0}}$ Thus, the period of a pendulum given an amplitude of 90° has a period about 18% longer than that given by the simple harmonic oscillator. Rewrite the period in terms of elliptic integrals. We first restate the integral to be evaluated. ${\displaystyle T=2{\sqrt {\frac {2l}{g}}}\int _{0}^{\theta _{0}}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta -\cos \theta _{0}}}}}$ Make use of the following substitutions. The third line immediately follows from the second substitution. ${\displaystyle \cos \theta =1-2\sin ^{2}{\frac {\theta }{2}}}$ ${\displaystyle \sin {\frac {\theta }{2}}=\sin {\frac {\theta _{0}}{2}}\sin \phi }$ ${\displaystyle {\frac {1}{2}}\cos {\frac {\pi }{2}}{\mathrm {d} }\theta =2k\cos \phi {\mathrm {d} }\phi }$ For simplicity, let ${\displaystyle k=\sin {\frac {\theta _{0}}{2}}.}$ Notice that when ${\displaystyle \theta =0,}$ ${\displaystyle \phi =0,}$ and when ${\displaystyle \theta =\theta _{0},}$ ${\displaystyle \phi =\pi /2.}$ ${\displaystyle {\begin{aligned}T&=2{\sqrt {\frac {2l}{g}}}\int _{0}^{\theta _{0}}{\frac {{\mathrm {d} }\theta }{\sqrt {\cos \theta -\cos \theta _{0}}}}\\&=2{\sqrt {\frac {l}{g}}}\int _{0}^{\theta _{0}}{\frac {{\mathrm {d} }\theta }{\sqrt {\sin ^{2}{\frac {\theta _{0}}{2}}-\sin ^{2}{\frac {\theta }{2}}}}}\\&=2{\sqrt {\frac {l}{g}}}\int _{0}^{\pi /2}{\frac {1}{k}}{\frac {2k{\mathrm {d} }\phi }{\sqrt {1-\sin ^{2}\phi }}}{\frac {\sqrt {1-\sin ^{2}\phi }}{\sqrt {1-\sin ^{2}{\frac {\theta }{2}}}}}\\&=4{\sqrt {\frac {l}{g}}}\int _{0}^{\pi /2}{\frac {{\mathrm {d} }\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}\end{aligned}}}$ This integral is called the complete elliptic integral of the first kind, denoted by ${\displaystyle K(k).}$ This integral does not have a solution expressible in terms of elementary functions, but it can be expressed as a series by way of the Beta function again. The period can thus be written exactly as follows. ${\displaystyle T=4{\sqrt {\frac {l}{g}}}K\left(\sin ^{2}{\frac {\theta _{0}}{2}}\right)}$ Evaluate the elliptic integral using the Beta function. A more detailed explanation of this evaluation can be found here. We must make use of the binomial series. ${\displaystyle (1-x)^{\alpha }=\sum _{m=0}^{\infty }{\alpha \choose m}(-1)^{m}x^{m}}$ ${\displaystyle {\begin{aligned}K(k)&=\int _{0}^{\pi /2}{\frac {{\mathrm {d} }\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}\\&=\int _{0}^{\pi /2}{\mathrm {d} }\phi \sum _{m=0}^{\infty }{-1/2 \choose m}(-1)^{m}k^{2m}\sin ^{2m}\phi \\&=\sum _{m=0}^{\infty }{-1/2 \choose m}(-1)^{m}k^{2m}\int _{0}^{\pi /2}\sin ^{2m}\phi {\mathrm {d} }\phi \\&=\sum _{m=0}^{\infty }{\frac {(-1)^{m}\Gamma (1/2)}{m!\Gamma (1/2-m)}}{\frac {\Gamma (1/2)\Gamma (1/2+m)}{2\,m!}}k^{2m}\\&={\frac {\pi }{2}}\sum _{m=0}^{\infty }{\frac {(-1)^{m}\Gamma (1/2+m)}{(m!)^{2}\Gamma (1/2-m)}}k^{2m}\\&={\frac {\pi }{2}}\sum _{m=0}^{\infty }{\frac {(-1)^{m}\Gamma ^{2}(1/2+m)}{\pi (m!)^{2}}}\sin(\pi (m+1/2))k^{2m}\\&={\frac {\pi }{2}}\sum _{m=0}^{\infty }\left[{\frac {(2m-1)!!}{2^{m}m!}}\right]^{2}k^{2m}\\&={\frac {\pi }{2}}\left[1+{\frac {1}{2^{2}}}k^{2}+\left({\frac {3\cdot 1}{2\cdot 2}}\right)^{2}{\frac {1}{2!^{2}}}k^{4}+\left({\frac {5\cdot 3\cdot 1}{2\cdot 2\cdot 2}}\right)^{2}{\frac {1}{3!^{2}}}k^{6}+\cdots \right]\end{aligned}}}$ In this derivation, we used the binomial series, the relation between the Gamma and the factorial functions ${\displaystyle \Gamma (z)=(z-1)!,}$ Euler's reflection formula to simplify the ${\displaystyle \Gamma (1/2+m)}$ and ${\displaystyle \Gamma (1/2-m)}$ terms, the fact that ${\displaystyle (-1)^{m}\sin(\pi (m+1/2))=1}$ for all integers ${\displaystyle m,}$ and the double factorial identity relating it to the Gamma function, written below. ${\displaystyle (2m-1)!!={\frac {2^{m}\Gamma (m+1/2)}{\sqrt {\pi }}}}$ Examine the series. This is a very important series, and from this, we obtain the period of a true pendulum. Let ${\displaystyle T_{0}=2\pi {\sqrt {\frac {l}{g}}}}$ be the period of the pendulum using the small-angle approximation. The series clearly demonstrates the deviation from this approximation as ${\displaystyle \theta _{0}}$ gets larger. Since the region of convergence is ${\displaystyle |k|<1,}$ we see that at 180°, the series diverges, corresponding to a pendulum at unstable equilibrium. Remember that ${\displaystyle k=\sin {\frac {\theta _{0}}{2}}}$ in this relation. ${\displaystyle T=T_{0}\left[1+{\frac {1}{2^{2}}}k^{2}+\left({\frac {3\cdot 1}{2\cdot 2}}\right)^{2}{\frac {1}{2!^{2}}}k^{4}+\left({\frac {5\cdot 3\cdot 1}{2\cdot 2\cdot 2}}\right)^{2}{\frac {1}{3!^{2}}}k^{6}+\cdots \right]}$ The graph above shows the elliptic integral in red, along with its series expansions truncated to 2nd, 4th, and 10th order. We can clearly see the divergence here, as well as the series being progressively better approximations the more terms we keep. EditRelated wikiHows Evaluate Complete Elliptic Integrals Integrate Using the Beta Function

You are subscribed to email updates from How to of the Day. To stop receiving these emails, you may unsubscribe now.	Email delivery powered by Google
Google Inc., 1600 Amphitheatre Parkway, Mountain View, CA 94043, United States