This is beyond me right now... 4. Extra Creait: 10 pts The period of a pendulum with length L that makes a maximum angle of 690 with the vertical is dx T/2 1-k2 sin (x) where k =sin (160) and g is...
Q12 (5 points) The period of a pendulum with length L that makes a maximum angle Bo with the vertical is 1/2 dc T= T=4 9 11 - k2 sinº where k = sino and g is the acceleration due to gravity. When , is small, the approximation T 27 is used. We will see in this 9 exercise, that this is in fact the first term in the series expansion for T and that using the first two terms...
The period T of a pendulum with length L meters that makes a maximum angle of θ0 with the vertical is The vertical is: T= 4\sqrt{\frac{L}{9}}\int _0^{\frac{\pi }{2}}\frac{dx}{\sqrt{1-k^2sin^2x}} where k=sin((1/2)θ0) and g=9.8 m/sec2 in the acceleration due to gravity. (a) Find the first four terms of a series expansion for T by expanding the integrand using the binomial series and integrating term by term (your answer will include L, g, k). You may use the following integration fact: The integration...
T = 4V The figure shows a pendulum with length L that makes a maximum angle @o with the vertical. Using Newton's Second Law, it can be shown that the period T (the time for one complete swing) is given by -TT/2 dx L go 1 - k2 sin2(x) where k = sin(100) and g is the acceleration due to gravity. If L = 2 m and 60 = 46°, use Simpson's Rule with n = 10 to find the...