Hockey Game Sound Intensity Ratio: 112 DB Vs 118 DB

Nov 6, 2025 by Admin 52 views

Hey guys! Ever wondered how much louder one sound is compared to another? Let's dive into a fascinating physics problem involving sound intensity at a hockey game. We’ll break down the question step-by-step so it’s super easy to understand. This problem involves comparing the sound intensity of two hockey games where the sound levels were measured at different decibel levels. The key here is to understand the relationship between decibels (dB) and sound intensity and how to use logarithms to solve this type of problem. So, grab your mental hockey stick, and let's get started!

Understanding the Problem

In this scenario, we're given that the loudest sound during a hockey game on one night measured 112 dB, and the next night, it measured 118 dB. The main question we want to answer is: What fraction of the sound intensity of the second game was the sound intensity of the first game? This isn't just about simple subtraction; we need to delve into the logarithmic nature of the decibel scale.

The decibel scale is used to measure sound intensity, and it's a logarithmic scale. This means that an increase of 10 dB corresponds to a tenfold increase in sound intensity. This logarithmic relationship is crucial for solving the problem. Let's jot down the key information we have:

Sound level on the first night: 112 dB
Sound level on the second night: 118 dB

Our mission is to find the ratio of the sound intensity of the first night (I₁) to the sound intensity of the second night (I₂). This involves using the formula that relates sound level in decibels to sound intensity. We need to manipulate this formula to isolate the ratio I₁/I₂. Now, let's move on to the formula and how to use it to solve this cool sound puzzle!

The Decibel Formula

The decibel scale isn't linear; it's logarithmic, which means that each increase of 10 dB represents a tenfold increase in sound intensity. The formula that connects sound level (L) in decibels to sound intensity (I) is:

${ L = 10 \log_{10}(\frac{I}{I_0}) }$

Where:

L is the sound level in decibels (dB)
I is the sound intensity
I₀ is the reference intensity, which is the threshold of human hearing (10⁻¹² W/m²)

This formula is super important because it helps us bridge the gap between the decibel measurements we have and the sound intensity ratio we're trying to find. The reference intensity (I₀) is a fixed value, representing the quietest sound a human can typically hear. This provides a baseline for comparing different sound intensities.

Now, let’s apply this formula to our hockey game scenario. We'll have two equations, one for each night, and then we’ll see how we can manipulate these equations to find the ratio of the sound intensities. Next up, we'll set up the equations for each night and show you how the magic happens!

Setting Up the Equations

Alright, let’s put our formula to work! We have two sound levels, one for each night of the hockey game. We’ll use the decibel formula to create an equation for each night. This will help us relate the sound intensities (I₁ and I₂) to their respective decibel levels (L₁ and L₂).

For the first night, the sound level (L₁) was 112 dB. Plugging this into the formula, we get:

${ 112 = 10 \log_{10}(\frac{I_1}{I_0}) }$

Similarly, for the second night, the sound level (L₂) was 118 dB. So, the equation for the second night is:

${ 118 = 10 \log_{10}(\frac{I_2}{I_0}) }$

Now we have two equations, each representing the sound level on a particular night. Our goal is to find the ratio I₁/I₂. To do this, we need to manipulate these equations. A clever way to do this is to isolate the logarithm terms and then use properties of logarithms to combine the equations. This will lead us to an expression involving the ratio I₁/I₂. In the next section, we’ll walk through the steps of isolating the logarithms and setting up the ratio.

Isolating the Logarithms

Okay, let's get those logarithms by themselves! We have two equations:

${ 112 = 10 \log_{10}(\frac{I_1}{I_0}) }$

${ 118 = 10 \log_{10}(\frac{I_2}{I_0}) }$

To isolate the logarithm in the first equation, we divide both sides by 10:

${ \frac{112}{10} = \log_{10}(\frac{I_1}{I_0}) }$

So, we get:

${ 11.2 = \log_{10}(\frac{I_1}{I_0}) }$

We do the same for the second equation, dividing both sides by 10:

${ \frac{118}{10} = \log_{10}(\frac{I_2}{I_0}) }$

Which gives us:

${ 11.8 = \log_{10}(\frac{I_2}{I_0}) }$

Now we have the logarithms isolated. The next step is to use these equations to find the ratio of the sound intensities. A smart move here is to subtract one equation from the other. This will help us eliminate the I₀ term and get us closer to finding I₁/I₂. Ready to see how it's done? Let's dive into the next section!

Finding the Difference and the Ratio

Alright, let's find the difference between our two logarithmic equations. We have:

${ 11.2 = \log_{10}(\frac{I_1}{I_0}) }$

${ 11.8 = \log_{10}(\frac{I_2}{I_0}) }$

Subtract the first equation from the second:

${ 11.8 - 11.2 = \log_{10}(\frac{I_2}{I_0}) - \log_{10}(\frac{I_1}{I_0}) }$

This simplifies to:

${ 0.6 = \log_{10}(\frac{I_2}{I_0}) - \log_{10}(\frac{I_1}{I_0}) }$

Now, remember the logarithm property that ${\log_b(A) - \log_b(B) = \log_b(\frac{A}{B})}$ ? We can use this property to combine the logarithms:

${ 0.6 = \log_{10}(\frac{I_2/I_0}{I_1/I_0}) }$

The I₀ terms cancel out, leaving us with:

${ 0.6 = \log_{10}(\frac{I_2}{I_1}) }$

We're getting closer! Now we need to get rid of the logarithm. To do this, we can use the inverse operation, which is raising 10 to the power of both sides:

${ 10^{0.6} = 10^{\log_{10}(\frac{I_2}{I_1})} }$

This simplifies to:

${ 10^{0.6} = \frac{I_2}{I_1} }$

We want the ratio I₁/I₂, so we take the reciprocal of both sides:

${ \frac{1}{10^{0.6}} = \frac{I_1}{I_2} }$

Now, we just need to calculate ${10^{0.6}}$ and then take its reciprocal to find the ratio I₁/I₂. Let’s crunch those numbers in the next section!

Calculating the Ratio

Alright, let's get to the final calculation! We've found that the ratio of the sound intensities is:

${ \frac{I_1}{I_2} = \frac{1}{10^{0.6}} }$

Now we need to calculate ${10^{0.6}}$ . Using a calculator, we find that:

${ 10^{0.6} \approx 3.981 }$

So, the ratio I₁/I₂ is:

${ \frac{I_1}{I_2} = \frac{1}{3.981} }$

${ \frac{I_1}{I_2} \approx 0.251 }$

This means that the sound intensity of the first game was approximately 0.251 times the sound intensity of the second game. To put it another way, the sound intensity on the first night was about 25.1% of the sound intensity on the second night. Cool, right?

Final Answer and Implications

So, we've crunched the numbers and found that the fraction of the sound intensity of the first game compared to the second game is approximately 0.251. That means the sound intensity at the first hockey game was about one-quarter of the sound intensity at the second game.

In Summary:

Sound level first night: 112 dB
Sound level second night: 118 dB
Ratio of sound intensities (I₁/I₂): 0.251

This problem highlights how the logarithmic decibel scale works. A difference of 6 dB (118 dB - 112 dB) might not seem like much, but it represents a significant difference in sound intensity. The second game was almost four times as intense in terms of sound energy compared to the first game!

Understanding these concepts is super useful, not just for physics class but also for understanding everyday phenomena. Sound levels can impact our hearing and overall health, so knowing how to interpret decibel measurements can help us make informed decisions about our environment.

Guys, thanks for joining me in solving this sound intensity puzzle! I hope you found it as fascinating as I did. Keep exploring the world of physics – it’s full of amazing stuff!