A particle of mass m is executing oscillation about the origin on the X-axis. Its potential energy i

Q: Why is the time period not independent of amplitude like in standard Simple Harmonic Motion?

In standard SHM, the restoring force is linear ($F = -kx$), which arises from a quadratic potential ($U = \frac{1}{2}kx^2$). This linearity ensures that as amplitude increases, the restoring force increases proportionally, keeping the time period constant. For a cubic potential ($U = k|x|^3$), the restoring force is non-linear ($F \propto -x^2$). The force increases much faster with displacement, causing the particle to speed up more at larger amplitudes, which decreases the time period.

Q: How did we get the dimensions of potential energy $U$ as $[ML^2T^{-2}]$?

Potential energy, like any form of energy or work, has the unit of Joules in SI. Since Work = Force $\times$ Displacement, its dimensions are $[\text{Force}] \times [\text{Length}] = [MLT^{-2}] \times [L] = [ML^2T^{-2}]$.

Q: Can we use this dimensional analysis method for any potential of the form $U(x) = k|x|^n$?

Yes! For any potential of the form $U(x) = k|x|^n$, you can show using dimensional analysis that the time period scales as $T \propto a^{\frac{2-n}{2n}}$. For $n=2$ (standard SHM), $T \propto a^0$ (independent of $a$). For $n=3$ (this problem), $T \propto a^{-1/2}$.

Q: What does the negative sign in the exponent of amplitude $a^{-1/2}$ physically mean?

A negative exponent means that the time period $T$ is inversely proportional to a power of the amplitude $a$. Physically, this means that if you pull the particle further away (larger amplitude), it will complete one full oscillation in less time because the restoring force at larger displacements is extremely strong.

Q: Is the motion of this particle still considered Simple Harmonic Motion?

No, the motion is periodic and oscillatory, but it is not 'Simple' Harmonic Motion. SHM strictly requires the restoring force to be directly proportional to the displacement ($F \propto -x$). Here, the restoring force is proportional to $-x^2$ (for $x > 0$), making it a non-linear oscillation.

Question

A particle of mass $m$ is executing oscillation about the origin on the $X$-axis. Its potential energy is $U(x) = k|x|^3$, where $k$ is a positive constant. If the amplitude of oscillation is $a$, then its time period $T$ is

Options:

(A) proportional to $1/\sqrt{a}$

(B) independent of $a$

(C) proportional to $\sqrt{a}$

(D) proportional to $a^{3/2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle of mass $m$ trapped in a symmetric potential well.

Unlike the standard simple harmonic oscillator where the potential energy is quadratic ($U(x) = \frac{1}{2}kx^2$), here we are dealing with a non-linear potential well given by:

$$U(x) = k|x|^3$$

This cubic dependence changes the entire dynamics of the system!

At the extreme positions, where the displacement reaches the amplitude $x = \pm a$, the particle momentarily comes to rest.

Therefore, its kinetic energy at these turning points is zero, and the total mechanical energy $E$ is entirely potential:

$$E = ka^3$$

As the particle oscillates back and forth between $-a$ and $+a$, energy continuously sloshes between kinetic and potential forms.

Our goal is to find how the time period $T$ of this non-linear oscillation scales with the amplitude $a$.

### The Dimensional Analysis Shortcut

Solving the exact equation of motion for a cubic potential involves complex elliptic integrals.

But as physics students, we have a superpower: **Dimensional Analysis**.

Let us assume that the time period $T$ depends on the mass $m$, the amplitude $a$, and the potential constant $k$ through a power-law relation:

$$T \propto m^x a^y k^z$$

To find the exponents $x$, $y$, and $z$, we must first determine the dimensions of each quantity.

Time period $T$ has the dimension of time:

$$[T] = [T^1]$$

Mass $m$ has the dimension of mass:

$$[m] = [M^1]$$

Amplitude $a$ has the dimension of length:

$$[a] = [L^1]$$

Now, what about the constant $k$? We can find its dimensions from the potential energy equation $U(x) = k|x|^3$:

$$[k] = \frac{[U]}{[x^3]}$$

Since potential energy $U$ has the dimensions of work/energy ($[ML^2T^{-2}]$) and $x$ is a length ($[L]$), we get:

$$[k] = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$$

### Solving the Dimensional Equations

Now, let's substitute these dimensions back into our proportionality equation:

$$[M^0 L^0 T^1] = [M]^x [L]^y [ML^{-1}T^{-2}]^z$$

Combining the powers of $M$, $L$, and $T$ on the right-hand side, we get:

$$[M^0 L^0 T^1] = [M^{x+z} L^{y-z} T^{-2z}]$$

For this equation to hold true, the powers of mass, length, and time on both sides must be identical. This gives us a system of three simple linear equations:

1. For Time ($T$):
   $$-2z = 1 \implies z = -\frac{1}{2}$$

2. For Length ($L$):
   $$y - z = 0 \implies y = z = -\frac{1}{2}$$

3. For Mass ($M$):
   $$x + z = 0 \implies x = -z = \frac{1}{2}$$

This is incredibly elegant!

We have found that the time period $T$ scales with the amplitude $a$ as:

$$T \propto a^y \implies T \propto a^{-1/2} = \frac{1}{\sqrt{a}}$$

Thus, the time period is inversely proportional to the square root of the amplitude.

As the amplitude of oscillation increases, the time period actually **decreases**!

This is a stark contrast to standard SHM, where the time period is completely independent of the amplitude.

### Verification via Rigorous Integration

If you are still skeptical, let us verify this result using the laws of mechanics.

From the conservation of mechanical energy, the sum of kinetic and potential energy at any point $x$ is equal to the total energy $E$:

$$\frac{1}{2}mv^2 + k|x|^3 = ka^3$$

Solving for the velocity $v = \frac{dx}{dt}$ in the region $x \ge 0$:

$$v = \frac{dx}{dt} = \sqrt{\frac{2k}{m}(a^3 - x^3)}$$

Separating variables to find the time $T$ for one complete oscillation (which is four times the time taken to go from $0$ to $a$):

$$T = 4 \int_0^a \frac{dx}{\sqrt{\frac{2k}{m}(a^3 - x^3)}}$$

To solve this integral, let us use the substitution $x = au$, which means $dx = a\,du$.

Substituting this in, the limits of integration change from $[0, a]$ to $[0, 1]$:

$$T = 4 \int_0^1 \frac{a\,du}{\sqrt{\frac{2k}{m}(a^3 - a^3 u^3)}}$$

Factorizing $a^3$ out of the square root in the denominator:

$$T = 4 \int_0^1 \frac{a\,du}{a^{3/2} \sqrt{\frac{2k}{m}(1 - u^3)}} = 4 \sqrt{\frac{m}{2k}} a^{-1/2} \int_0^1 \frac{du}{\sqrt{1 - u^3}}$$

Since the integral $\int_0^1 \frac{du}{\sqrt{1 - u^3}}$ is just a constant numerical value, we can write:

$$T \propto a^{-1/2} = \frac{1}{\sqrt{a}}$$

Both methods yield the exact same beautiful scaling law!

This confirms that **Option (a)** is the correct answer.

Animated Solution for Physics - Oscillations: A particle of mass $m$ is executing oscillation about the origin on the $X$ -axis. Its potential energy is $U (x) = k ∣ x ∣^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is $a$ , then its time period $T$ is

Select Answer:

Visualized Solution

Understanding the Potential Well

The Power of Dimensional Analysis

Finding Dimensions of $k$

Setting up the Dimensional Equation

Grouping the Dimensional Powers

Comparing Powers of Time ( $T$ )

Comparing Powers of Length ( $L$ )

Comparing Powers of Mass ( $M$ )

Proportionality of Time Period

Alternative Method: Integration

Solving the Integral

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Dimensional Analysis Shortcut

Solving the Dimensional Equations

Verification via Rigorous Integration

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