Question

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?

Answer 1

Missing the formula

Answer 2

t=2pi spt root L/32

Answer 3

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

Answer 4

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

\[\frac{dT}{dL} = \frac{\pi}{16}\]