91影视

The total energy of the system under SHM is given by

${\mathit{E}}_{\mathbf{t}\mathbf{o}\mathbf{t}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{A}}^{\mathbf{2}}$

Here given that the spring force constant is $\mathrm{k}=225\mathrm{N}{\mathrm{m}}^{-1}$.

From the figure period of SHM is T =2 s .

Initial displacement at t =0 s i.e. amplitude at t =0 s is$A_1=7\mathrm{cm}=7脳{10}^{-2}\mathrm{m}$

The displacement at t=1 s i.e. amplitude at$t=1$s is $A_1=6\mathrm{cm}=6脳{10}^{-2}\mathrm{m}$.

Final displacement at t=4 i.e. amplitude at$t=4s$ is$A_4=3\mathrm{cm}=3脳{10}^{-2}\mathrm{m}$

(a) As in SHM at the maximum displacement, the speed of the object is zero i.e. object is not moving at maximum displacement. So from figure the maximum displacement of the object is at time t=0 s, 1 s, 2 s, 3 s, and 4 s.

Hence at time

t=0 s,1 s, 2 s, 3 s, and 4 s

the mass is not moving.

(b) The total energy of the system under SHM is given by

$E_{tot}=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2$ 鈥︹赌�..(1)

Where m is the mass, v is the speed, k is the spring force constant, and A is the amplitude of SHM.

Since initially the speed of the mass is zero $v=0m{s}^{-1}$, hence from above equation initially the original energy contained by the system will be

$E_{t}ot=\frac{1}{2}k{A}_0^2$

$=\frac{1}{2}脳225N{m}^{-1}脳(7脳{10}^{-2}m{)}^2$

Hence, the total energy in the system is $E_{tot}=0.55J$.

(c) Since the speed of the mass at maximum displacement is zero $v=0m{s}^{-1}$, hence from equation (1), at maximum displacement at time t=1 s s the energy contained by the system will be

$$\begin{gathered} {\mathrm{E}}_1=\frac{1}{2}{\mathrm{kA}}_1^2 \\ =\frac{1}{2}脳225{\mathrm{Nm}}^{鈭�1}脳{\left(6脳{10}^{鈭�2}\mathrm{m}\right)}^2 \\ =0.405\mathrm{J} \end{gathered}$$ 鈥︹赌︹赌�.(3)

Finally, at maximum displacement at time t=4 S the energy contained by the system will be

$$\begin{gathered} {\mathrm{E}}_4=\frac{1}{2}{\mathrm{kA}}_4^2 \\ =\frac{1}{2}脳225{\mathrm{Nm}}^{鈭�1}脳{\left(3脳{10}^{鈭�2}\mathrm{m}\right)}^2 \\ =0.101\mathrm{J} \end{gathered}$$ 鈥︹赌︹赌︹赌�.(4)

Hence from equations (3) and (4), the energy lose by the system between t=1 s and t =4 s will be

$$\begin{gathered} 螖E=E_1-E_4 \\ =0.405J-0.101J \\ =0.304J \end{gathered}$$

This energy will be stored in the system as a potential energy of the system.