Question

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance x from the 50 cm mark. The period of oscillation is observed to be 4.8 s. Find the distance x. m

Answer 1

Answer:

x=0.01457 m

Explanation:

From parallel axis theorem

I=Icm+mh²

where h=x

The rotational inertia about its center of mass is

Icm=mL²/12

where L=1.0 m

Thus T=4.8s  we obtain

$T=2\pi \sqrt{\frac{mL^{2}/12+mx^{2} }{mgx} }\\ T=2\pi \sqrt{\frac{L^{2} }{12gx}+x/g }\\ T^{2}=4\pi^{2}(\frac{L^{2} }{12gx}+x/g )/x\\ T^{2}x=\frac{\pi^{2} L^{2} }{3g}+(\frac{4\pi^{2} }{g})x^{2}\\ 0=(\frac{4\pi^{2} }{g} )x^{2}-(T^{2} )x+(\frac{\pi^{2}L^{2} }{3g} )\\ 4.03x^{2}-23.04x+0.335=0$

After Solving this quadratic we get

x₁=5.702 m

x₂=0.01457 m

One of the solution is an impossible value for x (x=5.70m is greater than L)

So we choose the other one

x=0.01457 m