91影视

The time period $T$of the simple pendulum can be given by

$T=2蟺\sqrt{\frac{L}{g}}$

$g$is the acceleration due to gravity. The period of simple pendulum is depend upon the the length and magnitude of gravitational constant.

Principle of energy conservation: The sum of the initial energies of the system plus the work done by the external forces on the system is equal to the sum of the final energies of the system:

$E_i+W=E_f$

Gravitational potential: The gravitational potential energy of a system - Earth is$U_g=mgy$

in which m is the mass of the item, $g=9.8""N/kg$, and y is the placement of the item with appreciate to the 0 degree of gravitational capacity electricity (the foundation of the coordinate device in our choice).

4- Kinetic Energy: The kinetic electricity of an item is:

$K=1/2m{v}^2$

in which m is the item's mass and v is its pace relative to the selected coordinate device.

The period of oscillation of lamp is $T=5.5s$

胃=3.0鈭�The maximum angle of lamp from vertical is$胃={3.0}^{鈭�}$

a) The objective is to find out the length of the lamp chain

b) The objective is to determine the maximum sped of the lamp

The swinging lamp is acted as simple pendulum. The period of oscillation can be obtained from equation 1

$T=2蟺\sqrt{\frac{L}{g}}$

To find $L$

$L=\frac{T^{2}g}{4{蟺}^2}$

$L=\frac{(5.5\mathrm{s}{)}^2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)}{4{蟺}^2}$

$=7.5m$

Define the system as ground and oscillating lights. Call the initial state of the system when the lamp is at its greatest angle to the vertical and the final state when the lamp is at its lowest point. Determine the final state of the pendulum to be the state of zero potential energy. The lamp has zero initial velocity because it stops momentarily at the maximum angle and has a maximum speed at the lowest point.

Apply the principle of conservation of energy to the lamp of equation (2) between the initial and final states described:

where $K_i=0$because the lamp has zero initial velocity, $U_{g}i=mg螖y$as found from equation (3) where 螖y is the height of the lamp from the lowest point, $W=0$because 'there is no external force up. system, $K_f=\frac{1}{2}m{v}_{max}^2$as found from equation (4) and $U_{g}f=0$because we define the final state as the zero potential state:

$0+mg\mathrm{螖}y+0=\frac{1}{2}m{v}_{max}^2+0$

$\begin{array}{c}\mathrm{mng}鈦�\mathrm{螖}y=\frac{1}{2}鈫�{\mathrm{惫谈}}_{max}^2\end{array}$

$\mathrm{螖}y=L鈭�L\cos 鈦�胃$

$\begin{array}{r}g(L鈭�L\cos 鈦�胃)=\frac{1}{2}{v}_{max}^2\\ gL(1鈭�\cos 鈦�胃)=\frac{1}{2}{v}_{max}^2\end{array}$

To find

$\begin{array}{c}{v}_{max}=\sqrt{2gL(1鈭�\cos 鈦�胃)}\\ v_{max}=\sqrt{2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(7.5\mathrm{m}\right)\left(1鈭�\cos 鈦�{3.0}^{鈭�}\right)}\\ =0.45\mathrm{m}/\mathrm{s}\end{array}$