Question

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50.00 cm mark. (a) The period of oscillation is 2.50 s. Find d. (b) If you moved the pivot 5.00 cm closer to the 50.00 cm mark, what would the period of oscillation be

Answer 1

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

$I=I_{cm}+md^2$

For a meter stick mass m , the rotational inertia about it's center of mass

$I_{cm}-\dfrac{mL^2}{12}$

Where, L = 1 m

Put the value into the formula of time period

$T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}$

$T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}$

$T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})$

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

$(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0$

Put the value of T, L and g into the formula

$4.028d^2-6.25d+0.336=0$

$d = 0.056\ m, 1.496\ m$

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

$T=2\pi\sqrt{\dfrac{l}{g}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}$

$T=1.35\ sec$

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.