## Related questions with answers

The period of a pendulum is given by $T = 4 \sqrt { \frac { \ell } { g } } \int _ { 0 } ^ { \pi / 2 } \frac { d \theta } { \sqrt { 1 - k ^ { 2 } \sin ^ { 2 } \theta } } = 4 \sqrt { \frac { \ell } { g } } F ( k )$, where $\ell$ is the length of the pendulum, $g \approx 9.8 \mathrm { m } / \mathrm { s } ^ { 2 }$ is the acceleration due to gravity, k=sin $\left( \theta _ { 0 } / 2 \right)$, and $\theta _ { 0 }$ is the initial angular displacement of the pendulum (in radians). The integral in this formula F(k) is called an elliptic integral, and it cannot be evaluated analytically.

Approximate F(0.1) by expanding the integrand in a Taylor (binomial) series and integrating term by term.

Solution

VerifiedGiven is the following function for period:

$T = 4 \sqrt{ \dfrac{l}{g} } \int_0^{\pi/2} \dfrac{ d\theta }{ \sqrt{1-k^2 \sin^2 \theta} } = 4\sqrt{ \dfrac{l}{g} } F(k)$

and $k = \sin (\theta_0 / 2)$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Thomas' Calculus

14th Edition•ISBN: 9780134438986 (8 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir#### Calculus: Early Transcendentals

2nd Edition•ISBN: 9780321947345 (2 more)Bernard Gillett, Lyle Cochran, William L. Briggs#### Calculus: Early Transcendentals

8th Edition•ISBN: 9781285741550 (5 more)James Stewart#### Calculus: Early Transcendentals

9th Edition•ISBN: 9781337613927Daniel K. Clegg, James Stewart, Saleem Watson## More related questions

1/4

1/7