91影视

1. Mass of the block,$m=2.00\text{kg}$
2. Spring constant, $k=100\text{N}/\text{m}$
3. At localid="1662972921566" $t=0\text{s},$position $x=0.129\text{m}$, and velocity localid="1662972887523" $v=3.415\text{m}/\text{s}$.

Using the relation between angular velocity and spring constant, we can find the angular velocity. Then, using this angular velocity, position function, and velocity function, we can find the amplitude of the oscillations, position, and velocity of the block at $\mathbf{\text{t}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{\text{s}}$.

Formula:

Angular frequency of a body in oscillation:

$蝇=\sqrt{\frac{k}{m}}$ ......(i)

The general expression for the velocity of motion:

$x=x_m\cos \left(蝇t+蠒\right)$ ......(ii)

The general expression for the velocity of motion:

$v=-x_m蝇\text{sin}\left(蝇t+蠒\right)$ ......(iii)

Dividing equation (iii) by (ii), we get

$\begin{array}{rcl}\frac{v}{x}& =& \frac{x_m蝇\sin \left(蝇t+蠒\right)}{x_m\cos \left(蝇t+蠒\right)}\\ \frac{v}{x}& =& -蝇\tan \left(蝇t+蠒\right)\\ \left(蝇t+蠒\right)& =& {\tan }^{-1}\left(-\frac{v}{蝇x}\right)\\ & & \end{array}$

Using equation (i) and the given values, the angular frequency can be given as:

$\begin{array}{rcl}蝇& =& \sqrt{\frac{100\text{N}/\text{m}}{2.00\text{kg}}}\\ & =& 7.07\text{rad}/\text{s}\end{array}$

Again,

$\begin{array}{rcl}\left(蝇\text{t}+蠒\right)& =& {\text{tan}}^{-1}\left[-\frac{-3.415\frac{\text{m}}{\text{s}}}{\left(7.07\frac{\text{rad}}{\text{s}}\right)\left(0.129\text{m}\right)}\right]\\ & =& -1.31\text{rad}\end{array}$

Hence, using the above value in equation (ii) and $x=0.129\text{m}$, we get

$\begin{array}{rcl}0.129\text{m}& =& x_m\text{cos}\left(-1.31\right)\\ x_m& =& 0.500\text{m}\end{array}$

Therefore, the amplitude of the oscillations is $0.50\text{m}$

At $t=1.00\text{s}$and $蝇=7.07\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$using these values in equation (a), we get

$蠒=-7.38\text{rad}$

From equation (ii), the position can be found as

$\begin{array}{rcl}{x}_0& =& x_m\cos \left(蠒\right)\\ & =& 0.500\text{m}\text{cos}\left(-7.38\text{rad}\right)\\ & =& -0.228\text{m}\end{array}$

Therefore, the position of the block at $\text{t}=0\text{s}$ is $-0.228\text{m}$

Using equation (iii) and the values$t=0\text{s},蠒=-8.38\text{rad},x_m=0.500\text{m}$, we get,

$\begin{array}{rcl}{v}_0& =& \left(-0.500\text{m}\right)蝇\text{sin}\left(-7.38\text{rad}\right)\\ & =& 3.15\text{m}/\text{s}\end{array}$

Therefore, the velocity of the block at $t=0\text{s}$is $3.15\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.