A gas bubble, from an explosion under water, oscillates with a period T proportional to p^a d^b E^c,

Q: Why do we assume $T = k p^a d^b E^c$?

In dimensional analysis, we assume a product-type power dependence between the physical quantities. The dimensionless constant $k$ accounts for any numerical factors that dimensions cannot predict.

Q: What is the dimensional formula for pressure?

Pressure is Force/Area. Force is $[\text{MLT}^{-2}]$ and Area is $[\text{L}^2]$. Dividing them gives $[\text{ML}^{-1}\text{T}^{-2}]$.

Q: Why is the power of M zero on the LHS?

The LHS represents the time period $T$, which only has the dimension of time. Therefore, its mass and length dimensions are zero, written as $[\text{M}^0\text{L}^0\text{T}^1]$.

Q: Can dimensional analysis find the exact formula?

No, it cannot determine the value of the dimensionless constant $k$. It only gives the proportional dependence of the variables.

Q: What if there were four unknown variables instead of three?

With only three fundamental dimensions (M, L, T) in mechanics, we can only form three independent equations. If there are four unknowns, we cannot uniquely solve for all powers using dimensional analysis alone.

Question

A gas bubble, from an explosion under water, oscillates with a period $T$ proportional to $p^a d^b E^c$, where $p$ is the static pressure, $d$ is the density of water and $E$ is the total energy of the explosion. Find the values of $a, b$ and $c$.

Sigma Percentile · Accepted Answer

### The Power of Dimensional Analysis

Imagine a massive underwater explosion. A gas bubble forms and begins to oscillate violently. We are given that its time period, $T$, depends on the static pressure $p$, the density of water $d$, and the total energy $E$.

Our goal is to find the exact powers $a$, $b$, and $c$ using the principle of dimensional homogeneity. This principle states that for any physically meaningful equation, the dimensions on the left-hand side must perfectly match the dimensions on the right-hand side.

### Setting Up the Dimensions

To use this principle, we first need the dimensional formula for each physical quantity involved.

The time period is simply time, so its dimension is:
$$[T] = [	ext{M}^0 	ext{L}^0 	ext{T}^1]$$

Pressure is force per unit area. Force is mass times acceleration, so pressure gives us:
$$[p] = \left[\frac{	ext{MLT}^{-2}}{	ext{L}^2}ight] = [	ext{M L}^{-1} 	ext{T}^{-2}]$$

Density is mass per unit volume, which translates to:
$$[d] = \left[\frac{	ext{M}}{	ext{L}^3}ight] = [	ext{M L}^{-3} 	ext{T}^0]$$

And energy has the dimensions of work (force times displacement), which is:
$$[E] = [	ext{MLT}^{-2} \cdot 	ext{L}] = [	ext{M L}^2 	ext{T}^{-2}]$$

### The Master Equation

Now, let's substitute these dimensional formulas back into our assumed proportional relation, $T = k \cdot p^a d^b E^c$.

$$[	ext{M}^0 	ext{L}^0 	ext{T}^1] = [	ext{M L}^{-1} 	ext{T}^{-2}]^a [	ext{M L}^{-3}]^b [	ext{M L}^2 	ext{T}^{-2}]^c$$

Let's carefully combine the powers for each fundamental dimension on the right side.

For Mass (M):
$$	ext{M}^{a+b+c}$$

For Length (L):
$$	ext{L}^{-a-3b+2c}$$

For Time (T):
$$	ext{T}^{-2a-2c}$$

### Solving the System

By equating the powers of M, L, and T from both sides, we generate a system of three linear equations:

**1. Equating powers of M:**
$$a + b + c = 0$$

**2. Equating powers of L:**
$$-a - 3b + 2c = 0$$

**3. Equating powers of T:**
$$-2a - 2c = 1$$

Let's solve them. From the third equation, we can express $c$ in terms of $a$:
$$c = -\frac{1}{2} - a$$

Substituting this into the first equation beautifully cancels out $a$:
$$a + b + \left(-\frac{1}{2} - aight) = 0 \implies b = \frac{1}{2}$$

Then, plugging $b$ and $c$ into the second equation allows us to solve for $a$:
$$-a - 3\left(\frac{1}{2}ight) + 2\left(-\frac{1}{2} - aight) = 0$$
$$-3a - \frac{5}{2} = 0 \implies a = -\frac{5}{6}$$

With the value of $a$ in hand, we can quickly find $c$:
$$c = -\frac{1}{2} - \left(-\frac{5}{6}ight) = \frac{1}{3}$$

### The Final Result

So, our final powers are:
$$a = -\frac{5}{6}, \quad b = \frac{1}{2}, \quad c = \frac{1}{3}$$

This means the time period is proportional to $p^{-5/6}$, $d^{1/2}$, and $E^{1/3}$.

**Physical Interpretation:** A larger explosion energy $E$ means a longer oscillation period, while a higher surrounding pressure $p$ compresses the bubble faster, reducing the period. Dimensional analysis is truly a powerful tool to uncover these hidden physical relationships!

Visualized Solution

$T \propto p^{a} d^{b} E^{c}$

$Dimensional Formulae$

$Substituting Dimensions$

$Equating Powers of M$

$Equating Powers of L$

$Equating Powers of T$

$Solving the System$

$Final Values$

$Physical Interpretation$

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Setting Up the Dimensions

The Master Equation

Solving the System

The Final Result

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Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensional Formulae

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Dimensions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Powers of M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Powers of L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Powers of T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Interpretation