Question

A particle of mass 200 gm moves in a potential U. Particle has energy 40 J. Time period of oscillat particle if U is described as U = -2x for x < 0, U = 4 x 10-3.x² for x > 0

Answer 1

T=4+5π s (approximately 19.71 s)

Answer 2

Solution:

We are given a particle of mass

$m=200\,\text{gm}=0.2\,\text{kg}$

moving in a piecewise potential

$$U(x)=\begin{cases} -2x & \text{if } x < 0, \\ 0.004\,x^2 & \text{if } x>0, \end{cases}$$

with total energy $E=40\,\text{J}$.

Step 1. Determine turning points

1. For $x<0$:
   The turning point $x_L$ is found from:

   $$E=-2x_L \quad\Rightarrow\quad -2x_L=40 \quad\Rightarrow\quad x_L=-20.$$

2. For $x>0$:
   The turning point $x_R$ is found from:

   $$0.004\,x_R^2=40 \quad\Rightarrow\quad x_R^2=\frac{40}{0.004}=10000 \quad\Rightarrow\quad x_R=100.$$

Thus, the particle oscillates between $x=-20$ and $x=100$.

Step 2. Write the time period integral

The time $dt$ for an infinitesimal displacement $dx$ is given by:

$$dt=\frac{dx}{\sqrt{\frac{2}{m}\Bigl[E-U(x)\Bigr]}}.$$

Since the potential is piecewise, we break the time for a half-oscillation ($t_{1/2}$) into two parts:

$$t_{1/2} = t_1 + t_2,$$

where

• $t_1=$ time from $x=-20$ to $x=0$ (using $U=-2x$),
• $t_2=$ time from $x=0$ to $x=100$ (using $U=0.004\,x^2$).

Then the full period is:

$$T=2(t_1+t_2).$$

Calculation for $t_1$ (from $x=-20$ to $0$):

For $x<0,$ $U(x)=-2x$ so:

$$E-U = 40-(-2x)=40+2x.$$

Also,

$$\frac{2}{m}=\frac{2}{0.2}=10.$$

Thus,

$$t_1 = \int_{-20}^{0}\frac{dx}{\sqrt{10\,(40+2x)}}.$$

Factor $40+2x=2(20+x)$ and pull out constants:

$$t_1 = \frac{1}{\sqrt{10}}\int_{-20}^{0}\frac{dx}{\sqrt{2(20+x)}} = \frac{1}{\sqrt{20}} \int_{-20}^{0}\frac{dx}{\sqrt{20+x}}.$$

Let

$$u=20+x,\quad du=dx.$$

When $x=-20$, $u=0$; when $x=0$, $u=20$. Then,

$$t_1 = \frac{1}{\sqrt{20}} \int_{0}^{20}\frac{du}{\sqrt{u}} = \frac{1}{\sqrt{20}} \cdot \left[2\sqrt{u}\right]_0^{20} = \frac{2\sqrt{20}}{\sqrt{20}}=2\,\text{s}.$$

Calculation for $t_2$ (from $x=0$ to $100$):

For $x>0,$ $U(x)=0.004\,x^2$ so:

$$E-U=40-0.004\,x^2.$$

Then,

$$t_2 = \int_{0}^{100}\frac{dx}{\sqrt{10\,(40-0.004x^2)}}.$$

Write:

$$t_2 = \frac{1}{\sqrt{10}}\int_{0}^{100}\frac{dx}{\sqrt{40-0.004x^2}}.$$

Factor 40 from the square root:

$$\sqrt{40-0.004x^2}=\sqrt{40}\sqrt{1-\frac{0.004}{40}x^2}.$$

Note that

$$\frac{0.004}{40}=0.0001=\frac{1}{10000},$$

so,

$$\sqrt{40-0.004x^2}=\sqrt{40}\sqrt{1-\frac{x^2}{10000}}.$$

Then,

$$t_2 = \frac{1}{\sqrt{10}\sqrt{40}}\int_{0}^{100}\frac{dx}{\sqrt{1-\frac{x^2}{10000}}} = \frac{1}{\sqrt{400}}\int_{0}^{100}\frac{dx}{\sqrt{1-\frac{x^2}{10000}}}.$$

Since $\sqrt{400}=20$,

$$t_2 = \frac{1}{20}\int_{0}^{100}\frac{dx}{\sqrt{1-\frac{x^2}{10000}}}.$$

Now use the substitution:

$$\frac{x}{100}=\sin\theta,\quad dx=100\cos\theta\,d\theta.$$

When $x=0$, $\theta=0$; when $x=100$, $\theta=\frac{\pi}{2}$. Also,

$$\sqrt{1-\frac{x^2}{10000}} = \sqrt{1-\sin^2\theta} = \cos\theta.$$

Thus,

$$t_2 = \frac{1}{20}\int_{0}^{\frac{\pi}{2}}\frac{100\cos\theta\,d\theta}{\cos\theta} = \frac{100}{20}\int_{0}^{\frac{\pi}{2}}d\theta = 5\left(\frac{\pi}{2}\right)=\frac{5\pi}{2}\,\text{s}.$$

Step 3. Calculate the total period

The half period is:

$$t_{1/2}=t_1+t_2=2+\frac{5\pi}{2}.$$

Thus, the full period is:

$$T=2\,(t_1+t_2)=2\left(2+\frac{5\pi}{2}\right)=4+5\pi\,\text{s}.$$

Final Answer:

$$\boxed{T=4+5\pi\,\text{s} \quad\text{(approximately 19.71 s)}}$$