The pressure wave p = 0.01 sin[1000t - 3x] Nm^-2, corresponds to the sound produced by a vibrating b

Q: Why do we use $v = \frac{\omega}{k}$?

The wave speed $v$ is the distance a wave crest travels per unit time. Since a full wave cycle covers a distance of $\lambda$ in time $T$, $v = \frac{\lambda}{T}$. Multiplying numerator and denominator by $2\pi$ gives $v = \frac{2\pi/T}{2\pi/\lambda} = \frac{\omega}{k}$.

Q: Why must the temperature be in Kelvin?

The formula $v = \sqrt{\frac{\gamma RT}{M}}$ is derived from the ideal gas law, where $T$ represents absolute temperature. Using Celsius would imply zero speed at $0^\circ\text{C}$ and imaginary speeds at negative temperatures, which is physically incorrect!

Q: Does the pressure amplitude $0.01$ affect the speed?

No! For linear acoustic waves, the speed of sound depends entirely on the properties of the medium (temperature, molar mass, adiabatic index), not on the amplitude or frequency of the wave itself.

Q: What if the frequency of the blade changed?

Changing the frequency of the source changes the wavelength of the sound produced, but the speed of sound remains constant as long as the medium's properties (like temperature) don't change.

Q: How do we quickly square $1.008$ without a calculator?

You can use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$. Here, $(1 + 0.008)^2 \approx 1 + 2(0.008) = 1.016$. This makes manual calculations much faster and is highly recommended for JEE!

Question

The pressure wave $p = 0.01 \sin[1000t - 3x] 	ext{ Nm}^{-2}$, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is $0^\circ	ext{C}$. On some other day when temperature is $T$, the speed of sound produced by the same blade and at the same frequency is found to be $336 	ext{ ms}^{-1}$. Approximate value of $T$ is

Options:

(A) $15^\circ	ext{C}$

(B) $11^\circ	ext{C}$

(C) $12^\circ	ext{C}$

(D) $4^\circ	ext{C}$

Sigma Percentile · Accepted Answer

Imagine you are standing in a quiet room, and suddenly, a blade starts vibrating. What you hear as sound is actually a series of high and low-pressure regions traveling through the air. This is a longitudinal pressure wave!

In this problem, we are given the mathematical heartbeat of this wave:
$$p = 0.01 \sin(1000t - 3x)$$

I know this equation might look like a random jumble of numbers, but let's take a breath and decode it. Every single number here tells a physical story.

### The Anatomy of a Sound Wave

The general equation for a traveling pressure wave is:
$$p = p_0 \sin(\omega t - kx)$$

By simply comparing our given equation to this standard form, we can extract the DNA of our wave.

The term $p_0 = 0.01 	ext{ N/m}^2$ is the pressure amplitude. It tells us how loud the sound is. But for finding the speed, we don't care about the loudness!

What we really care about are the terms inside the sine function. The coefficient of time $t$ is the angular frequency, $\omega = 1000 	ext{ rad/s}$. This tells us how fast the blade is oscillating.

The coefficient of position $x$ is the wave number, $k = 3 	ext{ m}^{-1}$. This tells us how tightly packed the wave crests are in space.

### Extracting the Hidden Speed

Now, how do we find the speed of the wave from $\omega$ and $k$?

Think about it: speed is distance over time. A wave travels a distance of one wavelength $\lambda$ in one time period $T$. So, $v = \frac{\lambda}{T}$.

If we multiply the numerator and denominator by $2\pi$, we get:
$$v = \frac{2\pi/T}{2\pi/\lambda}$$

And since $\omega = \frac{2\pi}{T}$ and $k = \frac{2\pi}{\lambda}$, we arrive at the beautiful relation:
$$v = \frac{\omega}{k}$$

Let's substitute our values:
$$v_1 = \frac{1000}{3} 	ext{ m/s}$$

This is the speed of sound on the first day, when the temperature was $0^\circ	ext{C}$.

### The Thermodynamics of Sound

Now, the problem takes a twist. On another day, the speed of sound changes to $v_2 = 336 	ext{ m/s}$. Why did the speed change if it's the same blade?

This is where the physics gets incredibly elegant. The speed of sound in a gas doesn't depend on the source; it depends entirely on the medium!

According to Laplace's correction to Newton's formula, the speed of sound in an ideal gas is given by:
$$v = \sqrt{\frac{\gamma RT}{M}}$$

Here, $\gamma$ is the adiabatic index, $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.

Since the gas (air) remains the same, $\gamma$, $R$, and $M$ are all constants. This leaves us with a profound proportionality:
$$v \propto \sqrt{T}$$

The speed of sound is directly proportional to the square root of the absolute temperature!

**Watch out for the trap here!** You must always use the absolute temperature in Kelvin. Using Celsius will lead to a catastrophic silly mistake.

So, our Day 1 temperature is:
$$T_1 = 0^\circ	ext{C} = 273 	ext{ K}$$

### The Master Calculation

We can now set up a ratio to compare the two days:
$$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$

Let's substitute the values we know:
$$\frac{336}{\frac{1000}{3}} = \sqrt{\frac{T_2}{273}}$$

Simplifying the left side:
$$\frac{336 	imes 3}{1000} = \frac{1008}{1000} = 1.008$$

So, we have:
$$1.008 = \sqrt{\frac{T_2}{273}}$$

To get rid of the square root, we square both sides:
$$(1.008)^2 = \frac{T_2}{273}$$

Now, you might be tempted to reach for a calculator, but in JEE, we use smart approximations! Using the binomial expansion $(1+x)^n \approx 1+nx$ for small $x$:
$$(1 + 0.008)^2 \approx 1 + 2(0.008) = 1.016$$

So, the equation becomes:
$$1.016 = \frac{T_2}{273}$$

Multiplying both sides by $273$:
$$T_2 = 273 	imes 1.016$$
$$T_2 = 273 + 273(0.016)$$
$$T_2 = 273 + 4.368 = 277.368 	ext{ K}$$

### The Final Verdict

We have found the temperature on the second day in Kelvin. But the options are in Celsius!

To convert back, we simply subtract $273$:
$$T_2 = 277.368 - 273 = 4.368^\circ	ext{C}$$

Looking at our options, the closest approximate value is **$4^\circ	ext{C}$**.

And there you have it! By understanding the anatomy of a wave and the thermodynamics of the medium, we seamlessly connected an abstract mathematical equation to a real-world temperature change. Keep visualizing the physics, and the math will always follow!

Select Answer:

Visualized Solution

$p = p_{0} sin (ω t - k x)$

$v_{1} = \frac{ω}{k}$

\text{Day 2 Parameters}

$v \propto T$

\text{Substitution}

\text{Calculation}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Wave Equation and Wave Speed

The Anatomy of a Sound Wave

Extracting the Hidden Speed

The Thermodynamics of Sound

The Master Calculation

The Final Verdict

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