Problem 4.3.6. The exact formula for the period of a simple pendulum of length L is...

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Problem 4.3.6. The exact formula for the period of a simple pendulum of length L is r/2 T = 4√L/9 ™1² do 1 – k² sin² -X Φ where g is the acceleration due to gravity and k = sin 2, a being the initial angular displacement. Show that in the limit of very small a this reduces to the formula we all learned as children: T = 2√L/g. By expanding the integrand in a series in k² sin² o, develop the next correction. What is the fractional change in T due to this term for a = π/3? The integral in question is called an elliptic integral. But we do not have to know anything about this fancy object if we just want to get an answer that is numerically very good for small oscillations. In general we can deal with many difficult integrals this way. For example, we do not know how to integrate e-* between finite limits. However if we want a good approximation to fe dx, for small a, we simply expand the exponential and integrate term by term. 2 Problem 4.3.6. The exact formula for the period of a simple pendulum of length L is r/2 T = 4√L/9 ™1² do 1 – k² sin² -X Φ where g is the acceleration due to gravity and k = sin 2, a being the initial angular displacement. Show that in the limit of very small a this reduces to the formula we all learned as children: T = 2√L/g. By expanding the integrand in a series in k² sin² o, develop the next correction. What is the fractional change in T due to this term for a = π/3? The integral in question is called an elliptic integral. But we do not have to know anything about this fancy object if we just want to get an answer that is numerically very good for small oscillations. In general we can deal with many difficult integrals this way. For example, we do not know how to integrate e-* between finite limits. However if we want a good approximation to fe dx, for small a, we simply expand the exponential and integrate term by term. 2