91Ó°ÊÓ

(a)

The potential energy of a particle oscillating on a spring equals zero at an equilibrium length.

It is obvious that the potential energy is zero $20cm$.

Hence, the spring's equilibrium length is $20cm$.

(b)

At turning points, the particle's velocity reverses.

The elastic potential energy equals the entire energy of the particle.

$\frac{1}{2}k{A}^2$, with maximum amplitude $A$.

Here, $K$is the spring constant.

From the figure localid="1648562916662" $14.9,\text{at}x=14\mathrm{cm}$

At $x=26\mathrm{cm}$

Potential energy makes up the entire energy.

(c)

According to the conservation of energy principle, elastic potential energy is equal to kinetic energy.

$\frac{1}{2}k{A}^2=\frac{1}{2}m{V}_{max}^2$

From the figure Q 14.9, the equilibrium position of particle is at $20cm$. From the figure Q14.9. kinetic energy of the particle is $7j$.

When a particle is in equilibrium, its entire energy is in the form of kinetic energy.

Hence, $K{E}_{max}=7\mathrm{J}$.

As a result, the particle's maximal kinetic energy is $7j$.

From the law of conversation $PE=K{E}_{max}$

Substitute $\frac{1}{2}k{A}^2\text{for}PE\text{, and}7\mathrm{J}\text{for}K{E}_{max}\text{in}$

Rearrange equation for $k$

$k=\frac{2\left(7\mathrm{J}\right)}{A^2}$

Subsitute $20\mathrm{cm}\text{for}A\text{in}k=\frac{2\left(7\mathrm{J}\right)}{A^2}$

$k=\frac{2\left(7\mathrm{J}\right)}{(20\mathrm{cm}{)}^2}$

$=\frac{2\left(7\mathrm{J}\right)}{{\left(\left(20\mathrm{cm}\right)\left(\frac{{10}^{−2}\mathrm{m}}{1\mathrm{cm}}\right)\right)}^2}$

$=\frac{2\left(7\mathrm{J}\right)}{(0.2\mathrm{m}{)}^2}$

$=350\mathrm{N}/\mathrm{m}$

When the net energy is doubled, the amplitude of motion is.

$14\mathrm{J}=\frac{1}{2}k{A}^2$

Rearrange the equation in $14\mathrm{J}=\frac{1}{2}k{A}^2\text{for}A$

$A^2=\frac{2\left(14\mathrm{J}\right)}{k}$

$A=\sqrt{\frac{2\left(14\mathrm{J}\right)}{k}}$

Substitute $A=\sqrt{\frac{2\left(14\mathrm{J}\right)}{k}}$

$A=\sqrt{\frac{2\left(14\mathrm{J}\right)}{350\mathrm{N}/\mathrm{m}}}$

$=\left(0.28\mathrm{m}\right)\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)$

$=28$

Since, the equilibrium at $20cm$

Both turning points are symmetrically placed on either side of the equilibrium.

So, the turning points are $20+8=28\mathrm{cm}\text{and}20−8=12\mathrm{cm}$.