OpenStudy (anonymous):

The period T of oscillation (in seconds) of a simple pendulum of Length L (in feet) is given by . What is the rate of change of T with respect to L when L = 16 ft?

7 years ago
jimthompson5910 (jim_thompson5910):

Missing the formula

7 years ago
OpenStudy (anonymous):

t=2pi spt root L/32

7 years ago
jimthompson5910 (jim_thompson5910):

T = 2pi * sqrt(L/32) dT/dt = pi * sqrt(16L)/(16L) * (dL/dt) (dT/dt)*(dt/dL) = [pi * sqrt(16L)/(16L) * (dL/dt)]*(dt/dL) dT/dL = pi * sqrt(16L)/(16L) dT/dL = pi * sqrt(16(16))/(16(16)) dT/dL = pi*sqrt(256)/256 dT/dL = 16pi/256 dT/dL = pi/16

7 years ago
OpenStudy (anonymous):

\[T = 2\pi * \sqrt{\frac{L}{32}}\] \[\frac{dT}{dL} = \frac{\pi}{16}\]

7 years ago