91Ó°ÊÓ

1- Energy conservation principle: The sum of a system's beginning energies plus the work done on it by external forces equals the sum of the system's end energies:

E_i +W = E_f

2. Kinetic energy: An object's kinetic energy is:

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

when m is the organism's mass and v speed

3- Gravitational potential energy: An object's gravitational potential energy is defined as:

PE = mgh

where m, is the object's mass, and h is height.

4- Elastic Potential Energy: The elastic potential energy of a spring-like item with a spring constant k stretched or compressed x from its undisturbed position is:

$U_s=\frac{1}{2}k{x}^2$

5- Particle moving in a straight line with constant acceleration: If a particle moves in a horizontal path with constant acceleration $a_y$, its motion is described by the kinematics equation, from which the following equation is derived:

$v_f^2=v_i^2+2a\mathrm{Δ}y$

6- The quadratic formula gives the answer to a quadratic equation of the form$a{x}^2+bx+x=0$.

$x=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$

7- In simple harmonic motion, the rate of an oscillator is given by

$f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m}}$

It is important to note which $f$does not depend on the amplitude, but rather on the mass $m$and force constant $k$.

Given information:

• The block is thrown from a height $h=\left(3.0\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.03\mathrm{m}$above the spring's peak.

• The oscillation amplitude of the unit is: $A=\left(10\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.10\mathrm{m}$.

Info that is required

The oscillation frequency of the block must be determined.

Assign the block, the spring, and the earth to the system. Let the system's initial state be when the block first contacts the spring, and the system's ultimate state be when the block reaches its lowest position and the spring is entirely compressed. Assume that the system's equilibrium position is the zero-gravitational-energy configuration. Between the described starting and end states, apply the principle of energy conservation from Equation(1) to the system comprising the block, the spring, and the ground.

$K_i+U_{gi}+U_{si}+W=K_f+U_{gf}+U_{sf}$

where $K=\frac{1}{2}m{v}^2$as determined by Equation (2), v being the speed of the block after falling a distance h,$U_{gi}=mg∆y$as determined by Equation (3), y being the spacing compressed by the spring,$U_{si}=0$ because the season is not compressed, $W=0$because there are no external pressures on the system, $K_f=0$ because the block stops temporarily at the bottom rank, $U_{gf}=mg(-A)$as determined by Equation (3), and $U_{sf}=\frac{1}{2}k{(∆y+A)}^2$, $∆y+A$is the distance squeezed by the force from the optimum configuration, as determined by Equation (4).

$\begin{array}{c}\frac{1}{2}m{v}^2+mg\mathrm{Δ}y+0+0=0+mg(−A)+\frac{1}{2}k(\mathrm{Δ}y+A{)}^2\\ \frac{1}{2}m{v}^2+mg\mathrm{Δ}y=\frac{1}{2}k(\mathrm{Δ}y+A{)}^2−mgA\end{array}$

Let the initial state be when the block is initially lowered from rest, and the final state be when the block hits the spring for the first time. Model the block as a particle moving vertically with continuous acceleration (gravitational acceleration) and solve equation($5$) to obtain the block's final speed, with the positive direction being downward.

localid="1651161119528" $v^2={v_i}^2+2gh$

localid="1651161123988" $v^2=0+2gh$

localid="1651161128632" $v=\sqrt{2gh}$

The gravitational and spring forces on the block are equal at equilibrium; the gravitational force $mg$is equal to the spring pressure $k∆y$on the block. So

$mg=k∆y$

Equate $k$:

$k=\frac{mg}{∆y}$

v and k from equations (9) and (10) should be substituted into equation (6): we get

$\frac{1}{2}m\left(2gh\right)+mg\mathrm{Δ}y=\frac{mg}{2\mathrm{Δ}y}(\mathrm{Δ}y+A{)}^2−mgA$

$mgh+mg\mathrm{Δ}y=\frac{my}{2\mathrm{Δ}y}(\mathrm{Δ}y+A{)}^2−mgA$

Both sides are divided by $mg$:

$h+\mathrm{Δ}y=\frac{(\mathrm{Δ}y+A{)}^2}{2\mathrm{Δ}y}−A$

$h+\mathrm{Δ}y=\frac{(\mathrm{Δ}y{)}^2+A^2+2A\mathrm{Δ}y}{2\mathrm{Δ}y}−A$$h+\mathrm{Δ}y=\frac{(\mathrm{Δ}y{)}^2+A^2+2A\mathrm{Δ}y}{2\mathrm{Δ}y}−A$

localid="1651161241780" $2h\mathrm{Δ}y+2(\mathrm{Δ}y{)}^2=(\mathrm{Δ}y{)}^2+A^2+2A\mathrm{Δ}y−2A\mathrm{Δ}y$

localid="1651161272326" ${(∆y)}^2+2h∆y-A^2=0$

For localid="1651161276979" $a=1,b=2h,$and localid="1651161280947" $c=-A^2$, find the solution using the quadratic formula from equation (localid="1651161291481" $6$).

localid="1651161285627" $\mathrm{Δ}y=\frac{−2h+\sqrt{4h^2+4A^2}}{0}$

localid="1651161299103" $=−h+\sqrt{h^2+A^2}$

Because a negative sign for distance is unphysical. we used the positive root. Replace the numerical values with their equivalents.

localid="1651161305962" $\mathrm{Δ}y=−\left(0.03\mathrm{m}\right)+\sqrt{(0.03\mathrm{m}{)}^2+(0.10\mathrm{m}{)}^2}$

localid="1651161310888" $=0.0744m$

The frequency of the block's oscillation is then calculated using equation $7$$7$) ,

localid="1651161320086" $f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m}}$

Substitute localid="1651161331392" $k$in the equation (localid="1651161344172" $6$)

localid="1651161324936" $f=\frac{1}{2\mathrm{π}}\sqrt{\frac{mg}{m\mathrm{Δ}y}}$

localid="1651161349528" $=\frac{1}{2\mathrm{π}}\sqrt{\frac{g}{\mathrm{Δ}y}}$

Substitute numerical value:

localid="1651161354387" $f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{9.80\mathrm{m}/{\mathrm{s}}^2}{0.0744\mathrm{m}}}$

localid="1651161358791" $=1.8\mathrm{Hz}$