Q.The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = kx^{2}/2$, where $k$ is the force constant of the oscillator. For $k = 0.5\ \text{N m}^{-1}$, the graph of $V(x)$ versus $x$ is shown in Fig. 5.12. Show that a particle of total energy $1\ \text{J}$ moving under this potential must 'turn back' when it reaches $x = \pm 2\ \text{m}$.

Step 1 — Locate the point.

Total energy $E = \text{PE}+\text{KE}$. With $\text{PE}:\text{KE}=1:15$,

Since $\text{PE}=\tfrac{1}{2}kx^2$ and $E=\tfrac{1}{2}kA^2$,

Step 2 — Find the spring constant $k$. …

The key idea here is that when a force varies with position, you cannot simply multiply force by displacement — you must integrate. The work done by a variable force in one dimension is the area under the force vs. displacement curve. Since the force is linear in $x$, the area is a trapezoid, but integration gives the exact result.

Let’s work through it step by step.

1. Recall the definition of work for a variable force When the force $F$ depends on position $x$, the work done in moving from $x_1$ to $x_2$ is

Here $F(x) = 3 + 2x$ (in newtons), $x_1 = 1.5$ m, $x_2 = 3.5$ m.

2. Set up the integral

3. Integrate term by term The antiderivative of $3$ is $3x$, and the antiderivative of $2x$ is $x^2$ (since $\int 2x\,dx = x^2$). So

4. Evaluate at the limits At $x = 3.5$:

At $x = 1.5$:

5. Subtract to find the work

So the work done is $16$ joules. …

Concept & Intuition

In simple harmonic motion, total mechanical energy $E$ is constant and splits between kinetic energy $K$ and potential energy $U$. The ratio $U:K$ tells us how far the particle is from equilibrium: at the extremes ($x = \pm A$), all energy is potential; at the mean position ($x=0$), all energy is kinetic. For any intermediate position, the fraction of energy that is potential is proportional to $x^2$. So given a ratio, we can directly solve for $x$ in terms of amplitude $A$.

Step-by-step solution

1. Write the energy expressions For a particle in SHM with amplitude $A = 6\ \text{cm}$, spring constant $k$, and displacement $x$ from mean position:

(This uses $v^2 = \omega^2 (A^2 - x^2)$ and $K = \frac12 m v^2$, with $\omega^2 = k/m$.)

2. Set up the given ratio The problem states $U : K = 4 : 5$, so

Substitute the expressions:

3. Solve for $x^2$ Cross-multiply:

Bring terms together:

Hence

In SHM the kinetic energy at displacement $x$ is

$\text{KE} = \tfrac12 m\omega^2 (A^2 - x^2),\quad A = 10\text{ cm}.$ …

The key idea is that a block placed on a curved surface will start to slip when the component of gravity pulling it down the slope exceeds the maximum static friction. On any curve, the slope at a point is given by the derivative $dy/dx$, which equals $\tan\theta$, where $\theta$ is the angle the tangent makes with the horizontal. The condition for no slipping is simply $\tan\theta \le \mu_s$.

For the parabola $y = x^2/20$, the slope increases as you move away from the vertex (the lowest point). So the block can be placed safely up to the point where the slope is just steep enough to make it slip. Beyond that, friction can't hold it.

1. Find the slope of the parabola. Differentiate:

This is $\tan\theta$ at any point $(x, y)$.

2. Apply the no-slip condition. The block stays put if

With $\mu_s = 0.5$, we have

So the maximum $x$ coordinate is $5$ m (the parabola is symmetric, so we consider positive $x$).

3. Find the corresponding height $y$. Substitute $x = 5$ into the parabola equation: …

We start by recognizing that the hill is smooth, meaning no friction. That lets us use conservation of mechanical energy to find the speed at the end of the hill. After that, the object leaves horizontally and falls under gravity — a classic projectile motion problem. The key is to connect the two stages cleanly.

1. Find the speed at the end of the hill

The object starts from rest at height $H$. At the end of the hill, its height is $H/2$.

By conservation of energy:

Cancel $m$ and multiply both sides by 2:

2. Analyze the projectile motion

At the end of the hill, the object moves horizontally with speed $v = \sqrt{gH}$ from a height $y_0 = H/2$ above the ground.

We set up coordinates:

• Initial vertical position: $y_0 = H/2$
• Initial vertical velocity: $v_{y0} = 0$
• Acceleration: $a_y = -g$ (downward)

The time $t$ to fall to the ground ($y=0$) is found from:

The key idea is that in SHM, total mechanical energy $E$ is constant and equals the maximum potential energy (when displacement equals amplitude). At any other position, potential energy is $U = \frac12 k x^2$, and since $E = \frac12 k A^2$, we have $U = E \cdot \frac{x^2}{A^2}$. So if we find the displacement at $t=0$, we can directly get the potential energy as a fraction of $E$.

1. Identify amplitude and phase.

   The position is $x = 5 \cos\left(\omega t + \frac{\pi}{4}\right)$ cm.

   The amplitude is $A = 5$ cm. The phase constant is $\phi = \frac{\pi}{4}$.

2. Find displacement at $t=0$.

   At $t=0$:

3. Relate potential energy to total energy. Total mechanical energy $E = 100$ J is the energy at maximum displacement $x = A$:

Potential energy at any $x$ is $U = \frac12 k x^2$.

So

4. Plug in the numbers. …